Solution
Subtask 1: This is a telescoping series with a k = k {\displaystyle a_{k}={\sqrt {k}}} . Taking a look at the partial sums we get
∑ 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} }} diverges, the series diverges as well.
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 } (harmonic series), ( s n ) {\displaystyle (s_{n})} is not bounded from above/below. Hence, the series diverges.
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}}}
Obviously, this is a telescoping series with a k = 1 k 2 {\displaystyle a_{k}={\tfrac {1}{k^{2}}}} . We get:
∑ 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 ) ( k + 2 ) ] = lim n → ∞ [ ∑ k = 1 n − 1 1 k ( k + 1 ) − 1 1 ⋅ 2 − ∑ k = 1 n 1 k ( k + 1 ) ( k + 2 ) ] = Computation rules for limits lim n → ∞ ∑ k = 1 n − 1 1 k ( k + 1 ) − 1 1 ⋅ 2 − lim n → ∞ ∑ k = 1 n 1 k ( k + 1 ) ( k + 2 ) = ∑ k = 1 ∞ 1 k ( k + 1 ) ⏟ = 1 − 1 2 − ∑ k = 1 ∞ 1 k ( k + 1 ) ( k + 2 ) ⏟ = 1 4 see 4. = 1 − 1 2 − 1 4 = 1 4 {\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}}}