Solution
Subtask 1: This is a telescoping series with
a
k
=
k
{\displaystyle a_{k}={\sqrt {k}}}
∑
k
=
1
n
(
k
+
1
−
k
)
=
n
+
1
−
1
{\displaystyle \sum _{k=1}^{n}\left({\sqrt {k+1}}-{\sqrt {k}}\right)={\sqrt {n+1}}-{\sqrt {1}}}
As
(
a
k
)
k
∈
N
{\displaystyle (a_{k})_{k\in \mathbb {N} }}
Alternative solution: One can easily find a lower bound for the partial sum sequence:
s
n
=
∑
k
=
1
n
(
k
+
1
−
k
)
=
∑
k
=
1
n
(
k
+
1
−
k
)
(
k
+
1
+
k
)
k
+
1
+
k
=
∑
k
=
1
n
k
+
1
−
k
k
+
1
+
k
=
∑
k
=
1
n
1
k
+
1
+
k
≥
∑
k
=
1
n
1
k
+
1
+
k
+
1
=
∑
k
=
1
n
1
2
k
+
1
≥
∑
k
=
1
n
1
2
(
k
+
1
)
{\displaystyle {\begin{aligned}s_{n}&=\sum \limits _{k=1}^{n}\left({\sqrt {k+1}}-{\sqrt {k}}\right)\\&=\sum \limits _{k=1}^{n}{\frac {\left({\sqrt {k+1}}-{\sqrt {k}}\right)\left({\sqrt {k+1}}+{\sqrt {k}}\right)}{{\sqrt {k+1}}+{\sqrt {k}}}}\\&=\sum \limits _{k=1}^{n}{\frac {k+1-k}{{\sqrt {k+1}}+{\sqrt {k}}}}\\&=\sum \limits _{k=1}^{n}{\frac {1}{{\sqrt {k+1}}+{\sqrt {k}}}}\\&\geq \sum \limits _{k=1}^{n}{\frac {1}{{\sqrt {k+1}}+{\sqrt {k+1}}}}\\&=\sum \limits _{k=1}^{n}{\frac {1}{2{\sqrt {k+1}}}}\\&\geq \sum \limits _{k=1}^{n}{\frac {1}{2(k+1)}}\end{aligned}}}
As
∑
k
=
1
∞
1
2
(
k
+
1
)
=
∞
{\displaystyle \sum \limits _{k=1}^{\infty }{\frac {1}{2(k+1)}}=\infty }
(
s
n
)
{\displaystyle (s_{n})}
Subtask 2: We have
2
k
+
1
k
2
(
k
+
1
)
2
=
k
2
+
2
k
+
1
−
k
2
k
2
(
k
+
1
)
2
=
(
k
+
1
)
2
−
k
2
k
2
(
k
+
1
)
2
=
(
k
+
1
)
2
k
2
(
k
+
1
)
2
−
k
2
k
2
(
k
+
1
)
2
=
1
k
2
−
1
(
k
+
1
)
2
{\displaystyle {\begin{aligned}{\frac {2k+1}{k^{2}(k+1)^{2}}}&={\frac {k^{2}+2k+1-k^{2}}{k^{2}(k+1)^{2}}}\\[0.5em]&={\frac {(k+1)^{2}-k^{2}}{k^{2}(k+1)^{2}}}\\[0.5em]&={\frac {(k+1)^{2}}{k^{2}(k+1)^{2}}}-{\frac {k^{2}}{k^{2}(k+1)^{2}}}\\[0.5em]&={\frac {1}{k^{2}}}-{\frac {1}{(k+1)^{2}}}\end{aligned}}}
a
k
=
1
k
2
{\displaystyle a_{k}={\tfrac {1}{k^{2}}}}
∑
k
=
1
∞
2
k
+
1
k
2
(
k
+
1
)
2
=
Telescoping
sum
∑
k
=
1
∞
(
1
k
2
−
1
(
k
+
1
)
2
)
=
lim
n
→
∞
(
1
1
2
−
1
(
n
+
1
)
2
)
=
1
{\displaystyle \sum _{k=1}^{\infty }{\frac {2k+1}{k^{2}(k+1)^{2}}}{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\sum _{k=1}^{\infty }\left({\frac {1}{k^{2}}}-{\frac {1}{(k+1)^{2}}}\right)=\lim _{n\to \infty }\left({\frac {1}{1^{2}}}-{\frac {1}{(n+1)^{2}}}\right)=1}
Subtask 3: Take a look at the hint. We get
1
k
(
k
+
2
)
=
1
2
⋅
2
k
(
k
+
2
)
=
1
2
⋅
(
k
+
2
−
k
)
k
(
k
+
2
)
=
1
2
(
k
+
2
)
k
(
k
+
2
)
−
1
2
k
k
(
k
+
2
)
=
1
2
k
−
1
2
k
+
2
{\displaystyle {\begin{aligned}{\frac {1}{k(k+2)}}&={\frac {{\frac {1}{2}}\cdot 2}{k(k+2)}}\\[0.5em]&={\frac {{\frac {1}{2}}\cdot (k+2-k)}{k(k+2)}}\\[0.5em]&={\frac {{\frac {1}{2}}(k+2)}{k(k+2)}}-{\frac {{\frac {1}{2}}k}{k(k+2)}}\\[0.5em]&={\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+2}}\end{aligned}}}
This is a more generalized version of a telescoping sum. The first and last two summands do not cancel:
∑
k
=
1
∞
1
k
(
k
+
2
)
=
lim
n
→
∞
∑
k
=
1
n
1
k
(
k
+
2
)
=
lim
n
→
∞
∑
k
=
1
n
(
1
2
k
−
1
2
k
+
2
)
=
lim
n
→
∞
1
2
∑
k
=
1
n
(
1
k
−
1
k
+
2
)
=
lim
n
→
∞
1
2
[
(
1
1
−
1
3
)
+
(
1
2
−
1
4
)
+
(
1
3
−
1
5
)
+
…
+
(
1
n
−
2
−
1
n
)
+
(
1
n
−
1
−
1
n
+
1
)
+
(
1
n
−
1
n
+
2
)
]
=
Telescoping
sum
lim
n
→
∞
1
2
[
1
1
+
1
2
−
1
n
+
1
−
1
n
+
2
]
=
1
2
⋅
[
1
+
1
2
]
=
1
2
⋅
3
2
=
3
4
{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k(k+2)}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{k(k+2)}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}\left({\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+2}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\sum _{k=1}^{n}\left({\frac {1}{k}}-{\frac {1}{k+2}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\left[\left({\frac {1}{1}}-{\frac {1}{3}}\right)+\left({\frac {1}{2}}-{\frac {1}{4}}\right)+\left({\frac {1}{3}}-{\frac {1}{5}}\right)+\ldots +\left({\frac {1}{n-2}}-{\frac {1}{n}}\right)+\left({\frac {1}{n-1}}-{\frac {1}{n+1}}\right)+\left({\frac {1}{n}}-{\frac {1}{n+2}}\right)\right]\\&{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\lim \limits _{n\to \infty }{\frac {1}{2}}\left[{\frac {1}{1}}+{\frac {1}{2}}-{\frac {1}{n+1}}-{\frac {1}{n+2}}\right]\\[0.5em]&={\frac {1}{2}}\cdot \left[1+{\frac {1}{2}}\right]\\[0.5em]&={\frac {1}{2}}\cdot {\frac {3}{2}}\\[0.5em]&={\frac {3}{4}}\end{aligned}}}
Subtask 4: We have
1
4
k
2
+
4
k
−
3
=
1
(
2
k
−
1
)
(
2
k
+
3
)
=
1
4
⋅
4
(
2
k
−
1
)
(
2
k
+
3
)
=
1
4
⋅
(
2
k
+
4
−
2
k
)
(
2
k
−
1
)
(
2
k
+
3
)
=
1
4
⋅
(
2
k
+
3
−
2
k
+
1
)
(
2
k
−
1
)
(
2
k
+
3
)
=
1
4
⋅
(
2
k
+
3
(
2
k
−
1
)
(
2
k
+
3
)
−
2
k
−
1
(
2
k
−
1
)
(
2
k
+
3
)
)
=
1
4
⋅
(
1
2
k
−
1
−
1
2
k
+
3
)
{\displaystyle {\begin{aligned}{\frac {1}{4k^{2}+4k-3}}&={\frac {1}{(2k-1)(2k+3)}}\\[0.5em]&={\frac {{\frac {1}{4}}\cdot 4}{(2k-1)(2k+3)}}\\[0.5em]&={\frac {{\frac {1}{4}}\cdot (2k+4-2k)}{(2k-1)(2k+3)}}\\[0.5em]&={\frac {{\frac {1}{4}}\cdot (2k+3-2k+1)}{(2k-1)(2k+3)}}\\[0.5em]&={\frac {1}{4}}\cdot \left({\frac {2k+3}{(2k-1)(2k+3)}}-{\frac {2k-1}{(2k-1)(2k+3)}}\right)\\[0.5em]&={\frac {1}{4}}\cdot \left({\frac {1}{2k-1}}-{\frac {1}{2k+3}}\right)\end{aligned}}}
We get the following telescoping series:
∑
k
=
1
∞
1
4
k
2
+
4
k
−
3
=
lim
n
→
∞
∑
k
=
1
n
1
4
k
2
+
4
k
−
3
=
lim
n
→
∞
∑
k
=
1
n
1
4
⋅
(
1
2
k
−
1
−
1
2
k
+
3
)
=
lim
n
→
∞
1
4
⋅
∑
k
=
1
n
(
1
2
k
−
1
−
1
2
k
+
3
)
=
lim
n
→
∞
1
4
[
(
1
1
−
1
5
)
+
(
1
3
−
1
7
)
+
(
1
5
−
1
9
)
+
…
+
(
1
2
n
−
5
−
1
2
n
−
1
)
+
(
1
2
n
−
3
−
1
2
n
+
1
)
+
(
1
2
n
−
1
−
1
2
n
+
3
)
]
=
Telescoping
sum
lim
n
→
∞
1
4
[
1
1
+
1
3
−
1
2
n
+
1
−
1
2
n
+
3
]
=
1
4
⋅
[
1
+
1
3
]
=
1
4
⋅
4
3
=
1
3
{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{4k^{2}+4k-3}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{4k^{2}+4k-3}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{4}}\cdot \left({\frac {1}{2k-1}}-{\frac {1}{2k+3}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{4}}\cdot \sum _{k=1}^{n}\left({\frac {1}{2k-1}}-{\frac {1}{2k+3}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{4}}\left[\left({\frac {1}{1}}-{\frac {1}{5}}\right)+\left({\frac {1}{3}}-{\frac {1}{7}}\right)+\left({\frac {1}{5}}-{\frac {1}{9}}\right)+\ldots +\left({\frac {1}{2n-5}}-{\frac {1}{2n-1}}\right)+\left({\frac {1}{2n-3}}-{\frac {1}{2n+1}}\right)+\left({\frac {1}{2n-1}}-{\frac {1}{2n+3}}\right)\right]\\&{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\lim \limits _{n\to \infty }{\frac {1}{4}}\left[{\frac {1}{1}}+{\frac {1}{3}}-{\frac {1}{2n+1}}-{\frac {1}{2n+3}}\right]\\[0.5em]&={\frac {1}{4}}\cdot \left[1+{\frac {1}{3}}\right]\\[0.5em]&={\frac {1}{4}}\cdot {\frac {4}{3}}\\[0.5em]&={\frac {1}{3}}\end{aligned}}}
Subtask 5: Take a look at the hint! We get
1
k
(
k
+
1
)
(
k
+
2
)
=
1
2
k
−
1
k
+
1
+
1
2
k
+
2
=
1
2
k
−
1
2
k
+
1
−
1
2
k
+
1
+
1
2
k
+
2
=
1
2
k
−
1
2
k
+
1
−
(
1
2
k
+
1
−
1
2
k
+
2
)
{\displaystyle {\begin{aligned}{\frac {1}{k(k+1)(k+2)}}&={\frac {\frac {1}{2}}{k}}-{\frac {1}{k+1}}+{\frac {\frac {1}{2}}{k+2}}\\[0.5em]&={\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+1}}-{\frac {\frac {1}{2}}{k+1}}+{\frac {\frac {1}{2}}{k+2}}\\[0.5em]&={\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+1}}-({\frac {\frac {1}{2}}{k+1}}-{\frac {\frac {1}{2}}{k+2}})\\[0.5em]\end{aligned}}}
Hence, we can calculate the series using two telescoping series:
∑
k
=
1
∞
1
k
(
k
+
1
)
(
k
+
2
)
=
lim
n
→
∞
∑
k
=
1
n
1
k
(
k
+
1
)
(
k
+
2
)
=
lim
n
→
∞
∑
k
=
1
n
(
1
2
k
−
1
2
k
+
1
−
(
1
2
k
+
1
−
1
2
k
+
2
)
)
=
lim
n
→
∞
1
2
∑
k
=
1
n
(
1
k
−
1
k
+
1
)
−
1
2
∑
k
=
1
n
(
1
k
+
1
−
1
k
+
2
)
=
Telescoping
sum
lim
n
→
∞
1
2
[
(
1
1
−
1
n
+
1
)
−
(
1
2
−
1
n
+
2
)
]
=
lim
n
→
∞
1
2
[
1
2
−
1
n
+
1
+
1
n
+
2
]
=
1
2
⋅
[
1
2
−
0
+
0
]
=
1
2
⋅
1
2
=
1
4
{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k(k+1)(k+2)}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}\left({\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+1}}-({\frac {\frac {1}{2}}{k+1}}-{\frac {\frac {1}{2}}{k+2}})\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\sum _{k=1}^{n}\left({\frac {1}{k}}-{\frac {1}{k+1}}\right)-{\frac {1}{2}}\sum _{k=1}^{n}\left({\frac {1}{k+1}}-{\frac {1}{k+2}}\right)\\[0.5em]&{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\lim \limits _{n\to \infty }{\frac {1}{2}}\left[\left({\frac {1}{1}}-{\frac {1}{n+1}}\right)-\left({\frac {1}{2}}-{\frac {1}{n+2}}\right)\right]\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\left[{\frac {1}{2}}-{\frac {1}{n+1}}+{\frac {1}{n+2}}\right]\\[0.5em]&={\frac {1}{2}}\cdot \left[{\frac {1}{2}}-0+0\right]\\[0.5em]&={\frac {1}{2}}\cdot {\frac {1}{2}}\\[0.5em]&={\frac {1}{4}}\end{aligned}}}
Alternative solution: It holds that
1
k
(
k
+
1
)
(
k
+
2
)
=
1
2
⋅
2
k
(
k
+
1
)
(
k
+
2
)
=
1
2
⋅
(
k
+
2
−
k
)
k
(
k
+
1
)
(
k
+
2
)
=
1
2
(
k
+
2
)
k
(
k
+
1
)
(
k
+
2
)
−
1
2
k
k
(
k
+
1
)
(
k
+
2
)
=
1
2
k
(
k
+
1
)
−
1
2
(
k
+
1
)
(
k
+
2
)
{\displaystyle {\begin{aligned}{\frac {1}{k(k+1)(k+2)}}&={\frac {{\frac {1}{2}}\cdot 2}{k(k+1)(k+2)}}\\[0.5em]&={\frac {{\frac {1}{2}}\cdot (k+2-k)}{k(k+1)(k+2)}}\\[0.5em]&={\frac {{\frac {1}{2}}(k+2)}{k(k+1)(k+2)}}-{\frac {{\frac {1}{2}}k}{k(k+1)(k+2)}}\\[0.5em]&={\frac {\frac {1}{2}}{k(k+1)}}-{\frac {\frac {1}{2}}{(k+1)(k+2)}}\\[0.5em]\end{aligned}}}
Using the properties of telescoping series, we get:
∑
k
=
1
∞
1
k
(
k
+
1
)
(
k
+
2
)
=
lim
n
→
∞
∑
k
=
1
n
1
k
(
k
+
1
)
(
k
+
2
)
=
lim
n
→
∞
∑
k
=
1
n
(
1
2
k
(
k
+
1
)
−
1
2
(
k
+
1
)
(
k
+
2
)
)
=
lim
n
→
∞
1
2
∑
k
=
1
n
(
1
k
(
k
+
1
)
−
1
(
k
+
1
)
(
k
+
2
)
)
=
Telescoping
sum
lim
n
→
∞
1
2
[
1
1
⋅
2
−
1
(
n
+
1
)
(
n
+
2
)
]
=
1
2
⋅
[
1
2
−
0
]
=
1
2
⋅
1
2
=
1
4
{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k(k+1)(k+2)}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}\left({\frac {\frac {1}{2}}{k(k+1)}}-{\frac {\frac {1}{2}}{(k+1)(k+2)}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\sum _{k=1}^{n}\left({\frac {1}{k(k+1)}}-{\frac {1}{(k+1)(k+2)}}\right)\\[0.5em]&{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\lim \limits _{n\to \infty }{\frac {1}{2}}\left[{\frac {1}{1\cdot 2}}-{\frac {1}{(n+1)(n+2)}}\right]\\[0.5em]&={\frac {1}{2}}\cdot \left[{\frac {1}{2}}-0\right]\\[0.5em]&={\frac {1}{2}}\cdot {\frac {1}{2}}\\[0.5em]&={\frac {1}{4}}\end{aligned}}}
Solution 6: It holds that
k
−
1
k
(
k
+
1
)
(
k
+
2
)
=
k
k
(
k
+
1
)
(
k
+
2
)
−
1
k
(
k
+
1
)
(
k
+
2
)
=
1
(
k
+
1
)
(
k
+
2
)
−
1
k
(
k
+
1
)
(
k
+
2
)
{\displaystyle {\frac {k-1}{k(k+1)(k+2)}}={\frac {k}{k(k+1)(k+2)}}-{\frac {1}{k(k+1)(k+2)}}={\frac {1}{(k+1)(k+2)}}-{\frac {1}{k(k+1)(k+2)}}}
It follows that
∑
k
=
1
∞
k
−
1
k
(
k
+
1
)
(
k
+
2
)
=
lim
n
→
∞
∑
k
=
1
n
k
−
1
k
(
k
+
1
)
(
k
+
2
)
=
lim
n
→
∞
∑
k
=
1
n
(
1
(
k
+
1
)
(
k
+
2
)
−
1
k
(
k
+
1
)
(
k
+
2
)
)
=
Index
shift
lim
n
→
∞
[
∑
k
=
2
n
−
1
1
k
(
k
+
1
)
−
∑
k
=
1
n
1
k
(
k
+
1
)
(
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{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {k-1}{k(k+1)(k+2)}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {k-1}{k(k+1)(k+2)}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}\left({\frac {1}{(k+1)(k+2)}}-{\frac {1}{k(k+1)(k+2)}}\right)\\[0.5em]&{\underset {\text{shift}}{\overset {\text{Index}}{=}}}\lim \limits _{n\to \infty }\left[\sum _{k=2}^{n-1}{\frac {1}{k(k+1)}}-\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\right]\\[0.5em]&=\lim \limits _{n\to \infty }\left[\sum _{k=1}^{n-1}{\frac {1}{k(k+1)}}-{\frac {1}{1\cdot 2}}-\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\right]\\[0.5em]&{\underset {\text{for limits}}{\overset {\text{Computation rules}}{=}}}\lim \limits _{n\to \infty }\sum _{k=1}^{n-1}{\frac {1}{k(k+1)}}-{\frac {1}{1\cdot 2}}-\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\\[0.5em]&=\underbrace {\sum _{k=1}^{\infty }{\frac {1}{k(k+1)}}} _{=1}-{\frac {1}{2}}-\underbrace {\sum _{k=1}^{\infty }{\frac {1}{k(k+1)(k+2)}}} _{={\frac {1}{4}}{\text{ see }}4.}\\[0.5em]&=1-{\frac {1}{2}}-{\frac {1}{4}}\\[0.5em]&={\frac {1}{4}}\end{aligned}}}
![{\displaystyle {\begin{aligned}{\frac {2k+1}{k^{2}(k+1)^{2}}}&={\frac {k^{2}+2k+1-k^{2}}{k^{2}(k+1)^{2}}}\\[0.5em]&={\frac {(k+1)^{2}-k^{2}}{k^{2}(k+1)^{2}}}\\[0.5em]&={\frac {(k+1)^{2}}{k^{2}(k+1)^{2}}}-{\frac {k^{2}}{k^{2}(k+1)^{2}}}\\[0.5em]&={\frac {1}{k^{2}}}-{\frac {1}{(k+1)^{2}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d66606d4866d972597fd9085a0ee8f9ffa70d7a5)
![{\displaystyle {\begin{aligned}{\frac {1}{k(k+2)}}&={\frac {{\frac {1}{2}}\cdot 2}{k(k+2)}}\\[0.5em]&={\frac {{\frac {1}{2}}\cdot (k+2-k)}{k(k+2)}}\\[0.5em]&={\frac {{\frac {1}{2}}(k+2)}{k(k+2)}}-{\frac {{\frac {1}{2}}k}{k(k+2)}}\\[0.5em]&={\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+2}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8230dfc9bf437c8db136d9556c9cc41bc4b15173)
![{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k(k+2)}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{k(k+2)}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}\left({\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+2}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\sum _{k=1}^{n}\left({\frac {1}{k}}-{\frac {1}{k+2}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\left[\left({\frac {1}{1}}-{\frac {1}{3}}\right)+\left({\frac {1}{2}}-{\frac {1}{4}}\right)+\left({\frac {1}{3}}-{\frac {1}{5}}\right)+\ldots +\left({\frac {1}{n-2}}-{\frac {1}{n}}\right)+\left({\frac {1}{n-1}}-{\frac {1}{n+1}}\right)+\left({\frac {1}{n}}-{\frac {1}{n+2}}\right)\right]\\&{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\lim \limits _{n\to \infty }{\frac {1}{2}}\left[{\frac {1}{1}}+{\frac {1}{2}}-{\frac {1}{n+1}}-{\frac {1}{n+2}}\right]\\[0.5em]&={\frac {1}{2}}\cdot \left[1+{\frac {1}{2}}\right]\\[0.5em]&={\frac {1}{2}}\cdot {\frac {3}{2}}\\[0.5em]&={\frac {3}{4}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba9653d34396c0a945040c774fe8388e7bc48c1e)
![{\displaystyle {\begin{aligned}{\frac {1}{4k^{2}+4k-3}}&={\frac {1}{(2k-1)(2k+3)}}\\[0.5em]&={\frac {{\frac {1}{4}}\cdot 4}{(2k-1)(2k+3)}}\\[0.5em]&={\frac {{\frac {1}{4}}\cdot (2k+4-2k)}{(2k-1)(2k+3)}}\\[0.5em]&={\frac {{\frac {1}{4}}\cdot (2k+3-2k+1)}{(2k-1)(2k+3)}}\\[0.5em]&={\frac {1}{4}}\cdot \left({\frac {2k+3}{(2k-1)(2k+3)}}-{\frac {2k-1}{(2k-1)(2k+3)}}\right)\\[0.5em]&={\frac {1}{4}}\cdot \left({\frac {1}{2k-1}}-{\frac {1}{2k+3}}\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9d7de7a82368439b72417d50528e3b98b1e9b40)
![{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{4k^{2}+4k-3}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{4k^{2}+4k-3}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{4}}\cdot \left({\frac {1}{2k-1}}-{\frac {1}{2k+3}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{4}}\cdot \sum _{k=1}^{n}\left({\frac {1}{2k-1}}-{\frac {1}{2k+3}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{4}}\left[\left({\frac {1}{1}}-{\frac {1}{5}}\right)+\left({\frac {1}{3}}-{\frac {1}{7}}\right)+\left({\frac {1}{5}}-{\frac {1}{9}}\right)+\ldots +\left({\frac {1}{2n-5}}-{\frac {1}{2n-1}}\right)+\left({\frac {1}{2n-3}}-{\frac {1}{2n+1}}\right)+\left({\frac {1}{2n-1}}-{\frac {1}{2n+3}}\right)\right]\\&{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\lim \limits _{n\to \infty }{\frac {1}{4}}\left[{\frac {1}{1}}+{\frac {1}{3}}-{\frac {1}{2n+1}}-{\frac {1}{2n+3}}\right]\\[0.5em]&={\frac {1}{4}}\cdot \left[1+{\frac {1}{3}}\right]\\[0.5em]&={\frac {1}{4}}\cdot {\frac {4}{3}}\\[0.5em]&={\frac {1}{3}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bdf2402906f4cbfb6e1cdf908b72b8e0429349e)
![{\displaystyle {\begin{aligned}{\frac {1}{k(k+1)(k+2)}}&={\frac {\frac {1}{2}}{k}}-{\frac {1}{k+1}}+{\frac {\frac {1}{2}}{k+2}}\\[0.5em]&={\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+1}}-{\frac {\frac {1}{2}}{k+1}}+{\frac {\frac {1}{2}}{k+2}}\\[0.5em]&={\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+1}}-({\frac {\frac {1}{2}}{k+1}}-{\frac {\frac {1}{2}}{k+2}})\\[0.5em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/004e9c225c66322f4f1626ad073e47dc47737706)
![{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k(k+1)(k+2)}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}\left({\frac {\frac {1}{2}}{k}}-{\frac {\frac {1}{2}}{k+1}}-({\frac {\frac {1}{2}}{k+1}}-{\frac {\frac {1}{2}}{k+2}})\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\sum _{k=1}^{n}\left({\frac {1}{k}}-{\frac {1}{k+1}}\right)-{\frac {1}{2}}\sum _{k=1}^{n}\left({\frac {1}{k+1}}-{\frac {1}{k+2}}\right)\\[0.5em]&{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\lim \limits _{n\to \infty }{\frac {1}{2}}\left[\left({\frac {1}{1}}-{\frac {1}{n+1}}\right)-\left({\frac {1}{2}}-{\frac {1}{n+2}}\right)\right]\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\left[{\frac {1}{2}}-{\frac {1}{n+1}}+{\frac {1}{n+2}}\right]\\[0.5em]&={\frac {1}{2}}\cdot \left[{\frac {1}{2}}-0+0\right]\\[0.5em]&={\frac {1}{2}}\cdot {\frac {1}{2}}\\[0.5em]&={\frac {1}{4}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66ce3bb056558fbe7d4191b18ffd12b834321cc1)
![{\displaystyle {\begin{aligned}{\frac {1}{k(k+1)(k+2)}}&={\frac {{\frac {1}{2}}\cdot 2}{k(k+1)(k+2)}}\\[0.5em]&={\frac {{\frac {1}{2}}\cdot (k+2-k)}{k(k+1)(k+2)}}\\[0.5em]&={\frac {{\frac {1}{2}}(k+2)}{k(k+1)(k+2)}}-{\frac {{\frac {1}{2}}k}{k(k+1)(k+2)}}\\[0.5em]&={\frac {\frac {1}{2}}{k(k+1)}}-{\frac {\frac {1}{2}}{(k+1)(k+2)}}\\[0.5em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a669054392a53e5791ebf9d8e2468d845bdf2378)
![{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k(k+1)(k+2)}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}\left({\frac {\frac {1}{2}}{k(k+1)}}-{\frac {\frac {1}{2}}{(k+1)(k+2)}}\right)\\[0.5em]&=\lim \limits _{n\to \infty }{\frac {1}{2}}\sum _{k=1}^{n}\left({\frac {1}{k(k+1)}}-{\frac {1}{(k+1)(k+2)}}\right)\\[0.5em]&{\underset {\text{sum}}{\overset {\text{Telescoping}}{=}}}\lim \limits _{n\to \infty }{\frac {1}{2}}\left[{\frac {1}{1\cdot 2}}-{\frac {1}{(n+1)(n+2)}}\right]\\[0.5em]&={\frac {1}{2}}\cdot \left[{\frac {1}{2}}-0\right]\\[0.5em]&={\frac {1}{2}}\cdot {\frac {1}{2}}\\[0.5em]&={\frac {1}{4}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19cf47bcbb3facbcc7a0c004402eacb15e872e88)
![{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {k-1}{k(k+1)(k+2)}}&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {k-1}{k(k+1)(k+2)}}\\[0.5em]&=\lim \limits _{n\to \infty }\sum _{k=1}^{n}\left({\frac {1}{(k+1)(k+2)}}-{\frac {1}{k(k+1)(k+2)}}\right)\\[0.5em]&{\underset {\text{shift}}{\overset {\text{Index}}{=}}}\lim \limits _{n\to \infty }\left[\sum _{k=2}^{n-1}{\frac {1}{k(k+1)}}-\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\right]\\[0.5em]&=\lim \limits _{n\to \infty }\left[\sum _{k=1}^{n-1}{\frac {1}{k(k+1)}}-{\frac {1}{1\cdot 2}}-\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\right]\\[0.5em]&{\underset {\text{for limits}}{\overset {\text{Computation rules}}{=}}}\lim \limits _{n\to \infty }\sum _{k=1}^{n-1}{\frac {1}{k(k+1)}}-{\frac {1}{1\cdot 2}}-\lim \limits _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{k(k+1)(k+2)}}\\[0.5em]&=\underbrace {\sum _{k=1}^{\infty }{\frac {1}{k(k+1)}}} _{=1}-{\frac {1}{2}}-\underbrace {\sum _{k=1}^{\infty }{\frac {1}{k(k+1)(k+2)}}} _{={\frac {1}{4}}{\text{ see }}4.}\\[0.5em]&=1-{\frac {1}{2}}-{\frac {1}{4}}\\[0.5em]&={\frac {1}{4}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb3b0dd1c2aee88c44849d09c68c7be587e915c7)
![{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {3^{k}}{7^{k}}}&=\sum _{k=0}^{\infty }\left({\frac {3}{7}}\right)^{k}-\underbrace {\frac {3^{0}}{7^{0}}} _{=1}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Geometric series with }}q={\frac {3}{7}}\right.}\\[0.5em]&={\frac {1}{1-{\frac {3}{7}}}}-1\\[0.5em]&={\frac {1}{\frac {4}{7}}}-1\\[0.5em]&={\frac {7}{4}}-1\\[0.5em]&={\frac {3}{4}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3da60bfca180e42b2801d672c345ab79aa3d15e)
![{\displaystyle {\begin{aligned}\sum _{k=0}^{\infty }{\frac {4}{(-3)^{k}}}&=\sum _{k=0}^{\infty }4\cdot \left(-{\frac {1}{3}}\right)^{k}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Factor rule for convergent series}}\right.}\\[0.5em]&=4\cdot \sum _{k=0}^{\infty }\left(-{\frac {1}{3}}\right)^{k}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Geometric series with }}q=-{\frac {1}{3}}\right.}\\[0.5em]&=4\cdot {\frac {1}{1-\left(-{\frac {1}{3}}\right)}}\\[0.5em]&=4\cdot {\frac {1}{\frac {4}{3}}}\\[0.5em]&=4\cdot {\frac {3}{4}}\\[0.5em]&=3\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4936e34c1f685252be946b64e83a4cf13117c489)
![{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }\left({\frac {2^{k}-1}{3^{k}}}\right)&=\sum _{k=1}^{\infty }\left({\frac {2^{k}}{3^{k}}}-{\frac {1}{3^{k}}}\right)\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Sum formula for convergent series}}\right.}\\[0.5em]&=\sum _{k=1}^{\infty }\left({\frac {2}{3}}\right)^{k}-\sum _{k=1}^{\infty }\left({\frac {1}{3}}\right)^{k}\\[0.5em]&=\sum _{k=0}^{\infty }\left({\frac {2}{3}}\right)^{k}-\underbrace {\left({\frac {2}{3}}\right)^{0}} _{=1}-\sum _{k=0}^{\infty }\left({\frac {1}{3}}\right)^{k}+\underbrace {\left({\frac {1}{3}}\right)^{0}} _{=1}\\[0.5em]&=\sum _{k=0}^{\infty }\left({\frac {2}{3}}\right)^{k}-\sum _{k=0}^{\infty }\left({\frac {1}{3}}\right)^{k}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Formula for geometric series}}\right.}\\[0.5em]&={\frac {1}{1-{\frac {2}{3}}}}-{\frac {1}{1-{\frac {1}{3}}}}\\[0.5em]&={\frac {1}{\frac {1}{3}}}-{\frac {1}{\frac {2}{3}}}\\[0.5em]&=3-{\frac {3}{2}}\\[0.5em]&={\frac {3}{2}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e709316476a7d9a28c322e661ebb564642f2a097)
![{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }a_{k}&=\sum _{k=0}^{\infty }\left({\frac {1}{2^{2k}}}+{\frac {1}{3^{2k+1}}}\right)\\[0.5em]&=\sum _{k=0}^{\infty }\left(\left({\frac {1}{4}}\right)^{k}+{\frac {1}{3}}\left({\frac {1}{9}}\right)^{k}\right)\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Sum formula and factor rule for convergent series}}\right.}\\[0.5em]&=\sum _{k=0}^{\infty }\left({\frac {1}{4}}\right)^{k}+{\frac {1}{3}}\cdot \sum _{k=0}^{\infty }\left({\frac {1}{9}}\right)^{k}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Formula for geometric series}}\right.}\\[0.5em]&={\frac {1}{1-{\frac {1}{4}}}}+{\frac {1}{3}}\cdot {\frac {1}{1-{\frac {1}{9}}}}\\[0.5em]&={\frac {1}{\frac {3}{4}}}+{\frac {1}{3}}\cdot {\frac {1}{\frac {8}{9}}}\\[0.5em]&={\frac {4}{3}}+{\frac {1}{3}}\cdot {\frac {9}{8}}\\[0.5em]&={\frac {4}{3}}+{\frac {3}{8}}\\[0.5em]&={\frac {32}{24}}+{\frac {9}{24}}\\[0.5em]&={\frac {41}{24}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37af866dcfd2fbec607d51f096c70a1388e2db06)
![{\displaystyle {\begin{aligned}\sum _{k=2}^{\infty }\left({\frac {3\cdot 2^{k+1}+9\cdot (-1)^{k}}{4\cdot 5^{k}}}\right)&=\sum _{k=0}^{\infty }\left({\frac {3\cdot 2\cdot 2^{k}+9\cdot (-1)^{k}}{4\cdot 5^{k}}}\right)-{\frac {3\cdot 2+9\cdot 1}{4\cdot 1}}-{\frac {3\cdot 4-9\cdot 1}{4\cdot 5}}\\[0.5em]&=\sum _{k=0}^{\infty }\left({\frac {6}{4}}\cdot {\frac {2^{k}}{5^{k}}}+{\frac {9}{4}}\cdot {\frac {(-1)^{k}}{5^{k}}}\right)-{\frac {15}{4}}-{\frac {3}{20}}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Sum formula and factor rule for convergent series}}\right.}\\[0.5em]&={\frac {6}{4}}\cdot \sum _{k=0}^{\infty }\left({\frac {2}{5}}\right)^{k}+{\frac {9}{4}}\cdot \sum _{k=0}^{\infty }\left(-{\frac {1}{5}}\right)^{k}-{\frac {15}{4}}-{\frac {3}{20}}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ {\text{Formula for geometric series}}\right.}\\[0.5em]&={\frac {3}{2}}\cdot {\frac {1}{1-{\frac {2}{5}}}}+{\frac {9}{4}}\cdot {\frac {1}{1+{\frac {1}{5}}}}-{\frac {15}{4}}-{\frac {3}{20}}\\[0.5em]&={\frac {3}{2}}\cdot {\frac {5}{3}}+{\frac {9}{4}}\cdot {\frac {5}{6}}-{\frac {75}{20}}-{\frac {3}{20}}\\[0.5em]&={\frac {5}{2}}+{\frac {15}{8}}-{\frac {78}{20}}\\[0.5em]&={\frac {20}{8}}+{\frac {15}{8}}-{\frac {78}{20}}\\[0.5em]&={\frac {35}{8}}-{\frac {78}{20}}\\[0.5em]&={\frac {175}{40}}-{\frac {156}{40}}\\[0.5em]&={\frac {19}{40}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4bc710d961d8389b451fd84c35e1f3bc5c701e0a)
![{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}&=\sum _{k=1}^{\infty }\left({\frac {1}{(2k-1)^{2}}}+{\frac {1}{(2k)^{2}}}\right)\\[0.5em]&\left\downarrow \ {\text{both series converge}}\right.\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {1}{(2k-1)^{2}}}+\sum _{k=1}^{\infty }{\frac {1}{(2k)^{2}}}\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {1}{(2k-1)^{2}}}+{\frac {1}{4}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72281dda0c355db8a6593c50d762187452c5f701)
![{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{(2k-1)^{2}}}&=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}-{\frac {1}{4}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\\[0.5em]&={\frac {3}{4}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\\[0.5em]&={\frac {3}{4}}\cdot {\frac {\pi ^{2}}{6}}\\[0.5em]&={\frac {\pi ^{2}}{8}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37e7dce138da56498eecd65c37ef05abd4ff3ef1)
![{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k^{2}}}&=\sum _{k=1}^{\infty }\left({\frac {1}{(2k-1)^{2}}}-{\frac {1}{(2k)^{2}}}\right)\\[0.5em]&\left\downarrow \ {\text{both series converge}}\right.\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {1}{(2k-1)^{2}}}-{\frac {1}{4}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\\[0.5em]&={\frac {3}{4}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}-{\frac {1}{4}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\\[0.5em]&={\frac {1}{2}}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}\\[0.5em]&={\frac {1}{2}}\cdot {\frac {\pi ^{2}}{6}}\\[0.5em]&={\frac {\pi ^{2}}{12}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/04421cc38291f57a3fd312484f580e86af002064)
![{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k(k+1)}}&=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k(k+1)}}\\[0.5em]&\left\downarrow \ {\text{Partial fraction decomposition}}\right.\\[0.5em]&=\sum _{k=1}^{\infty }(-1)^{k+1}\left({\frac {1}{k}}-{\frac {1}{k+1}}\right)\\[0.5em]&\left\downarrow \ {\text{series }}\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}{\text{ and }}\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k+1}}{\text{ converge}}\right.\\[0.5em]&=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}-\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k+1}}\\[0.5em]&=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}+\sum _{k=1}^{\infty }(-1)^{k+2}{\frac {1}{k+1}}\\[0.5em]&=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}+\sum _{k=0}^{\infty }(-1)^{k+2}{\frac {1}{k+1}}-1\\[0.5em]&\left\downarrow \ {\text{Index shift (2nd series)}}\right.\\[0.5em]&=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}+\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}-1\\[0.5em]&=\ln(2)+\ln(2)-1\\[0.5em]&=2\ln(2)-1\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0aafaa2cb3919ba548a0e3f0e5e7cfa4381f5c8f)
![{\displaystyle {\begin{aligned}\sum _{k=0}^{n}{\frac {k-1}{k!}}&\leq \sum _{k=0}^{n}{\frac {k}{k!}}\\[0.5em]&=\sum _{k=1}^{n}{\frac {k}{k!}}\\[0.5em]&=\sum _{k=1}^{n}{\frac {1}{(k-1)!}}\\[0.5em]&\left\downarrow \ {\text{Index shift}}\right.\\[0.5em]&=\sum _{k=0}^{n-1}{\frac {1}{k!}}\\[0.5em]&\leq \sum _{k=0}^{\infty }{\frac {1}{k!}}\\[0.5em]&=e<\infty \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36139ff8cffecb393d252e11cdb3e5d09fc82fbd)
![{\displaystyle {\begin{aligned}\sum _{k=0}^{\infty }{\frac {k-1}{k!}}&=\sum _{k=0}^{\infty }\left({\frac {k}{k!}}-{\frac {1}{k!}}\right)\\[0.5em]&\left\downarrow \ {\text{series converge}}\right.\\[0.5em]&=\sum _{k=0}^{\infty }{\frac {k}{k!}}-\sum _{k=0}^{\infty }{\frac {1}{k!}}\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {k}{k!}}-e\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {1}{(k-1)!}}-e\\[0.5em]&\left\downarrow \ {\text{Index shift}}\right.\\[0.5em]&=\sum _{k=0}^{\infty }{\frac {1}{k!}}-e\\[0.5em]&=e-e\\[0.5em]&=0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34472e85074e83d570276690fe28614c190186e3)
![{\displaystyle {\begin{aligned}\sum _{k=0}^{\infty }{\frac {k-1}{k!}}&=-1+\lim _{n\to \infty }\sum _{k=1}^{n}\left({\frac {k}{k!}}-{\frac {1}{k!}}\right)\\[0.5em]&=-1+\lim _{n\to \infty }\sum _{k=1}^{n}\left({\frac {1}{(k-1)!}}-{\frac {1}{k!}}\right)\\[0.5em]&\left\downarrow \ {\text{Telescoping sum}}\right.\\[0.5em]&=-1+\lim _{n\to \infty }\left({\frac {1}{0!}}-{\frac {1}{n!}}\right)\\[0.5em]&=-1+\left(1-\lim _{n\to \infty }{\frac {1}{n!}}\right)\\[0.5em]&=-1+1-0\\[0.5em]&=0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6fc2c790d36bdab53fe0123f93aab3b91c86b382)
![{\displaystyle {\begin{aligned}\sum _{k=0}^{n}{\frac {k+1}{k!}}&\leq \sum _{k=0}^{n}{\frac {k+k}{k!}}\\[0.5em]&=2\sum _{k=1}^{n}{\frac {k}{k!}}\\[0.5em]&=2\sum _{k=1}^{n}{\frac {1}{(k-1)!}}\\[0.5em]&\left\downarrow \ {\text{Index shift}}\right.\\[0.5em]&=2\sum _{k=0}^{n-1}{\frac {1}{k!}}\\[0.5em]&\leq 2\sum _{k=0}^{\infty }{\frac {1}{k!}}\\[0.5em]&=2e<\infty \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f563489011abe567650c236ad7dac887845b93cd)
![{\displaystyle {\begin{aligned}\sum _{k=0}^{\infty }{\frac {k+1}{k!}}&=\sum _{k=0}^{\infty }\left({\frac {k}{k!}}+{\frac {1}{k!}}\right)\\[0.5em]&\left\downarrow \ {\text{series converge}}\right.\\[0.5em]&=\sum _{k=0}^{\infty }{\frac {k}{k!}}+\sum _{k=0}^{\infty }{\frac {1}{k!}}\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {k}{k!}}+e\\[0.5em]&=\sum _{k=1}^{\infty }{\frac {1}{(k-1)!}}+e\\[0.5em]&\left\downarrow \ {\text{Index shift}}\right.\\[0.5em]&=\sum _{k=0}^{\infty }{\frac {1}{k!}}+e\\[0.5em]&=e+e\\[0.5em]&=2e\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cef323fbfb3e438329addfb16770feba2e141e43)
