Convergence of the exponential series
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At first, we need to show that the exponential series converges at all:
Theorem (convergence of the exponential series)
The series ∑ k = 0 ∞ 1 k ! {\displaystyle \sum _{k=0}^{\infty }{\tfrac {1}{k!}}} converges.
Proof (convergence of the exponential series)
We need to prove that the partial sums ( s n ) n ∈ N = ( ∑ k = 0 n 1 k ! ) n ∈ N {\displaystyle (s_{n})_{n\in \mathbb {N} }=\left(\sum _{k=0}^{n}{\tfrac {1}{k!}}\right)_{n\in \mathbb {N} }} converges. This can be done using the monotony criterion for sequences, where the partial sums ( s n ) {\displaystyle (s_{n})} form the sequence which we want to investigate.
Monotony is easy to see. Series elements are positive, so
s n + 1 = ∑ k = 0 n + 1 1 k ! = ∑ k = 0 n 1 k ! + 1 ( n + 1 ) ! ⏟ ≥ 0 ≥ ∑ k = 0 n 1 k ! = s n {\displaystyle s_{n+1}=\sum _{k=0}^{n+1}{\frac {1}{k!}}=\sum _{k=0}^{n}{\frac {1}{k!}}+\underbrace {\frac {1}{(n+1)!}} _{\geq 0}\geq \sum _{k=0}^{n}{\frac {1}{k!}}=s_{n}}
Hence, ( s n ) {\displaystyle (s_{n})} is monotonously increasing.
Boundedness from above can be shown by comparison to a geometric series with q = 1 2 < 1 {\displaystyle q={\tfrac {1}{2}}<1} . For partial sums, we have
∑ k = 0 n 1 k ! = ∑ k = 0 n 1 1 ⋅ 2 ⋅ 3 ⋅ … ⋅ ( k − 1 ) ⋅ k = ∑ k = 0 n 1 2 ⋅ 3 ⋅ … ⋅ ( k − 1 ) ⋅ k ↓ 1 2 , 1 3 , … 1 k − 1 , 1 k ≤ 1 2 ≤ ∑ k = 0 n 1 2 ⋅ 2 ⋅ … ⋅ 2 ⋅ 2 ⏟ ( k − 1 ) times = ∑ k = 0 n ( 1 2 ) k − 1 = 2 ⋅ ∑ k = 0 n ( 1 2 ) k ≤ 2 ⋅ ∑ k = 0 ∞ ( 1 2 ) k ↓ geometric series with q = 1 2 = 2 ⋅ 1 1 − 1 2 = 2 ⋅ 2 = 4 {\displaystyle {\begin{aligned}\sum _{k=0}^{n}{\frac {1}{k!}}&=\sum _{k=0}^{n}{\frac {1}{1\cdot 2\cdot 3\cdot \ldots \cdot (k-1)\cdot k}}\\[0.5em]&=\sum _{k=0}^{n}{\frac {1}{2\cdot 3\cdot \ldots \cdot (k-1)\cdot k}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\frac {1}{2}},{\frac {1}{3}},\ldots {\frac {1}{k-1}},{\frac {1}{k}}\leq {\frac {1}{2}}}\right.\\[0.5em]&\leq \sum _{k=0}^{n}\underbrace {\frac {1}{2\cdot 2\cdot \ldots \cdot 2\cdot 2}} _{(k-1){\text{ times}}}\\[0.5em]&=\sum _{k=0}^{n}\left({\frac {1}{2}}\right)^{k-1}\\[0.5em]&=2\cdot \sum _{k=0}^{n}\left({\frac {1}{2}}\right)^{k}\\[0.5em]&\leq 2\cdot \sum _{k=0}^{\infty }\left({\frac {1}{2}}\right)^{k}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{geometric series with }}q={\frac {1}{2}}}\right.\\[0.5em]&=2\cdot {\frac {1}{1-{\frac {1}{2}}}}\\[0.5em]&=2\cdot 2=4\end{aligned}}}
Hence, ( s n ) {\displaystyle (s_{n})} is bounded from above by 4 {\displaystyle 4} . So the monotony criterion implies convergence.
Limit of the exponential series
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Now, we show that the exponential series indeed converges towards Euler's number e {\displaystyle e} . This is done using the squeeze theorem by "squeezing" the partial sums ( s n ) n ∈ N = ( ∑ k = 0 n 1 k ! ) n ∈ N {\displaystyle (s_{n})_{n\in \mathbb {N} }=\left(\sum _{k=0}^{n}{\tfrac {1}{k!}}\right)_{n\in \mathbb {N} }} between the sequences ( ( 1 + 1 n ) n ) n ∈ N {\displaystyle ((1+{\tfrac {1}{n}})^{n})_{n\in \mathbb {N} }} and ( ( 1 + 1 n ) n + 1 ) n ∈ N {\displaystyle ((1+{\tfrac {1}{n}})^{n+1})_{n\in \mathbb {N} }} . Both bounding sequences converge to e {\displaystyle e} , so we get the desired result.
That means, we need to show:
( 1 + 1 n ) n ≤ ∑ k = 0 n 1 k ! ≤ ( 1 + 1 n ) n + 1 {\displaystyle (1+{\tfrac {1}{n}})^{n}\leq \sum _{k=0}^{n}{\frac {1}{k!}}\leq (1+{\tfrac {1}{n}})^{n+1}}
Theorem (limit of the exponential series)
There is ∑ k = 0 ∞ 1 k ! = e {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k!}}=e} .
Proof (limit of the exponential series)
We show
( 1 + 1 n ) n ≤ ∑ k = 0 n 1 k ! ≤ ( 1 + 1 n ) n + 1 {\displaystyle (1+{\tfrac {1}{n}})^{n}\leq \sum _{k=0}^{n}{\frac {1}{k!}}\leq (1+{\tfrac {1}{n}})^{n+1}}
and use the
squeeze theorem :
1st inequality: ( 1 + 1 n ) n ≤ ∑ k = 0 n 1 k ! {\displaystyle (1+{\tfrac {1}{n}})^{n}\leq \sum _{k=0}^{n}{\tfrac {1}{k!}}} . This is easier to establish than the second one. We need the binomial theorem ( 1 + x ) n = ∑ k = 0 n ( n k ) x k {\displaystyle (1+x)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{k}} with x = 1 n {\displaystyle x={\tfrac {1}{n}}} .
( 1 + 1 n ) n = binomial theorem ∑ k = 0 n ( n k ) 1 n k = ∑ k = 0 n n ⋅ ( n − 1 ) ⋅ ( n − 2 ) ⋅ … ⋅ ( n − k + 1 ) k ! ⋅ 1 n k = ∑ k = 0 n n ⋅ ( n − 1 ) ⋅ ( n − 2 ) ⋅ … ⋅ ( n − k + 1 ) n ⋅ n ⋅ n ⋅ … ⋅ n ⋅ 1 k ! = ∑ k = 0 n n n ⋅ n − 1 n ⋅ n − 2 n ⋅ … ⋅ n − k + 1 n ⏟ ≤ 1 ⋅ 1 k ! ≤ ∑ k = 0 n 1 k ! {\displaystyle {\begin{aligned}\left(1+{\frac {1}{n}}\right)^{n}&{\underset {\text{theorem}}{\overset {\text{binomial}}{=}}}\sum _{k=0}^{n}{\binom {n}{k}}{\frac {1}{n^{k}}}\\[0.5em]&=\sum _{k=0}^{n}{\frac {n\cdot (n-1)\cdot (n-2)\cdot \ldots \cdot (n-k+1)}{k!}}\cdot {\frac {1}{n^{k}}}\\[0.5em]&=\sum _{k=0}^{n}{\frac {n\cdot (n-1)\cdot (n-2)\cdot \ldots \cdot (n-k+1)}{n\cdot n\cdot n\cdot \ldots \cdot n}}\cdot {\frac {1}{k!}}\\[0.5em]&=\sum _{k=0}^{n}\underbrace {{\frac {n}{n}}\cdot {\frac {n-1}{n}}\cdot {\frac {n-2}{n}}\cdot \ldots \cdot {\frac {n-k+1}{n}}} _{\leq 1}\cdot {\frac {1}{k!}}\\[0.5em]&\leq \sum _{k=0}^{n}{\frac {1}{k!}}\end{aligned}}}
2nd inequality: ( 1 + 1 n ) n + 1 ≥ ∑ k = 0 n 1 k ! {\displaystyle (1+{\tfrac {1}{n}})^{n+1}\geq \sum _{k=0}^{n}{\tfrac {1}{k!}}} . Her, we additionally need the Bernoulli inequality ( 1 + x ) n ≥ 1 + n x {\displaystyle (1+x)^{n}\geq 1+nx} for x ≥ − 1 {\displaystyle x\geq -1} . In addition, a telescoping sum will appear in the end of the proof.
( 1 + 1 n ) n + 1 = binomial theorem ∑ k = 0 n + 1 ( n + 1 k ) 1 n k = 1 + ∑ k = 1 n + 1 ( n + 1 k ) 1 n k ↓ index shift = 1 + ∑ k = 0 n ( n + 1 k + 1 ) 1 n k + 1 = 1 + ∑ k = 0 n ( n + 1 ) ⋅ n ⋅ ( n − 1 ) ⋅ … ⋅ ( n − k + 1 ) ( k + 1 ) ! ⋅ 1 n k + 1 = 1 + ∑ k = 0 n ( n + 1 ) ⋅ n ⋅ ( n − 1 ) ⋅ … ⋅ ( n − ( k − 1 ) ) n ⋅ n ⋅ n ⋅ … ⋅ n ⋅ 1 ( k + 1 ) ! ↓ n + 1 , n , n − 1 , … n − ( k − 1 ) ≥ n − ( k − 1 ) ≥ 1 + ∑ k = 0 n ( n − ( k − 1 ) ) k + 1 n k + 1 ⋅ 1 ( k + 1 ) ! = 1 + ∑ k = 0 n ( 1 − k − 1 n ) k + 1 ⋅ 1 ( k + 1 ) ! ↓ Bernoulli's inequality with x = − k − 1 n ≥ − 1 ≥ 1 + ∑ k = 0 n ( 1 − ( k + 1 ) ( k − 1 ) n ) ⋅ 1 ( k + 1 ) ! ↓ Linearity of the sum = 1 + ∑ k = 0 n 1 ( k + 1 ) ! − 1 n ∑ k = 0 n ( k + 1 ) ( k − 1 ) ( k + 1 ) ! ↓ Index shift (1st sum), cancellation (2nd sum) = 1 + ∑ k = 1 n + 1 1 k ! − 1 n ∑ k = 0 n ( k − 1 ) k ! ↓ 1 = 1 1 ! and factor out the n + 1 -th summand of the sum for k = 0 and k = 1 for the 2nd sum. = ∑ k = 0 n 1 k ! + 1 ( n + 1 ) ! ⏟ ≥ 0 − 1 n [ − 1 + 0 + ∑ k = 2 n ( k − 1 ) k ! ] ↓ re-formulate the fraction within the 2nd sum and use k k ! = 1 ( k − 1 ) ! ≥ ∑ k = 0 n 1 k ! − 1 n [ − 1 + ∑ k = 2 n ( 1 ( k − 1 ) ! − 1 k ! ) ] ↓ telescoping sum ∑ k = 2 n ( 1 ( k − 1 ) ! − 1 k ! ) = 1 − 1 n ! = ∑ k = 0 n 1 k ! − 1 n [ − 1 + 1 − 1 n ! ] = ∑ k = 0 n 1 k ! + 1 n ⋅ n ! ⏟ ≥ 0 ≥ ∑ k = 0 n 1 k ! {\displaystyle {\begin{aligned}\left(1+{\frac {1}{n}}\right)^{n+1}&{\underset {\text{theorem}}{\overset {\text{binomial}}{=}}}\sum _{k=0}^{n+1}{\binom {n+1}{k}}{\frac {1}{n^{k}}}\\[0.5em]&=1+\sum _{k=1}^{n+1}{\binom {n+1}{k}}{\frac {1}{n^{k}}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{index shift}}}\right.\\[0.5em]&=1+\sum _{k=0}^{n}{\binom {n+1}{k+1}}{\frac {1}{n^{k+1}}}\\[0.5em]&=1+\sum _{k=0}^{n}{\frac {(n+1)\cdot n\cdot (n-1)\cdot \ldots \cdot (n-k+1)}{(k+1)!}}\cdot {\frac {1}{n^{k+1}}}\\[0.5em]&=1+\sum _{k=0}^{n}{\frac {(n+1)\cdot n\cdot (n-1)\cdot \ldots \cdot (n-(k-1))}{n\cdot n\cdot n\cdot \ldots \cdot n}}\cdot {\frac {1}{(k+1)!}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}n+1,n,n-1,\ldots n-(k-1)\geq n-(k-1)}\right.\\[0.5em]&\geq 1+\sum _{k=0}^{n}{\frac {(n-(k-1))^{k+1}}{n^{k+1}}}\cdot {\frac {1}{(k+1)!}}\\[0.5em]&=1+\sum _{k=0}^{n}\left(1-{\frac {k-1}{n}}\right)^{k+1}\cdot {\frac {1}{(k+1)!}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{Bernoulli's inequality with }}x=-{\frac {k-1}{n}}\geq -1}\right.\\[0.5em]&\geq 1+\sum _{k=0}^{n}\left(1-{\frac {(k+1)(k-1)}{n}}\right)\cdot {\frac {1}{(k+1)!}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{Linearity of the sum}}}\right.\\[0.5em]&=1+\sum _{k=0}^{n}{\frac {1}{(k+1)!}}-{\frac {1}{n}}\sum _{k=0}^{n}{\frac {(k+1)(k-1)}{(k+1)!}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{Index shift (1st sum), cancellation (2nd sum)}}}\right.\\[0.5em]&=1+\sum _{k=1}^{n+1}{\frac {1}{k!}}-{\frac {1}{n}}\sum _{k=0}^{n}{\frac {(k-1)}{k!}}\\[0.5em]&\left\downarrow \ {\color {OliveGreen}1={\frac {1}{1!}}{\text{ and factor out the }}n+1{\text{-th summand of the sum for }}k=0{\text{ and }}k=1{\text{ for the 2nd sum.}}}\right.\\[0.5em]&=\sum _{k=0}^{n}{\frac {1}{k!}}+\underbrace {\frac {1}{(n+1)!}} _{\geq 0}-{\frac {1}{n}}\left[-1+0+\sum _{k=2}^{n}{\frac {(k-1)}{k!}}\right]\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{re-formulate the fraction within the 2nd sum and use }}{\frac {k}{k!}}={\frac {1}{(k-1)!}}}\right.\\[0.5em]&\geq \sum _{k=0}^{n}{\frac {1}{k!}}-{\frac {1}{n}}\left[-1+\sum _{k=2}^{n}\left({\frac {1}{(k-1)!}}-{\frac {1}{k!}}\right)\right]\\[0.5em]&\left\downarrow \ {\color {OliveGreen}{\text{telescoping sum}}\sum _{k=2}^{n}\left({\frac {1}{(k-1)!}}-{\frac {1}{k!}}\right)=1-{\frac {1}{n!}}}\right.\\[0.5em]&=\sum _{k=0}^{n}{\frac {1}{k!}}-{\frac {1}{n}}\left[-1+1-{\frac {1}{n!}}\right]\\[0.5em]&=\sum _{k=0}^{n}{\frac {1}{k!}}+\underbrace {\frac {1}{n\cdot n!}} _{\geq 0}\\[0.5em]&\geq \sum _{k=0}^{n}{\frac {1}{k!}}\end{aligned}}}
In addition, we established ( 1 + 1 n ) n ≤ ∑ k = 0 n 1 k ! ≤ ( 1 + 1 n ) n + 1 {\displaystyle (1+{\tfrac {1}{n}})^{n}\leq \sum _{k=0}^{n}{\frac {1}{k!}}\leq (1+{\tfrac {1}{n}})^{n+1}} . Since lim n → ∞ ( 1 + 1 n ) n = lim n → ∞ ( 1 + 1 n ) n + 1 = e {\displaystyle \lim _{n\to \infty }(1+{\tfrac {1}{n}})^{n}=\lim _{n\to \infty }(1+{\tfrac {1}{n}})^{n+1}=e} , the squeeze theorem implies lim n → ∞ ∑ k = 0 n 1 k ! = ∑ k = 0 ∞ 1 k ! = e {\displaystyle \lim _{n\to \infty }\sum _{k=0}^{n}{\tfrac {1}{k!}}=\sum _{k=0}^{\infty }{\tfrac {1}{k!}}=e} .
Alternatively, one may show lim inf n → ∞ ∑ k = 0 n 1 k ! ≤ e ≤ lim sup n → ∞ ∑ k = 0 n 1 k ! {\displaystyle \liminf _{n\to \infty }\sum _{k=0}^{n}{\tfrac {1}{k!}}\leq e\leq \limsup _{n\to \infty }\sum _{k=0}^{n}{\tfrac {1}{k!}}} , which also implies ∑ k = 0 ∞ 1 k ! = e {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k!}}=e} .
Further, the sequences a n = ∑ k = 0 n 1 k ! {\displaystyle a_{n}=\sum _{k=0}^{n}{\tfrac {1}{k!}}} and b n = ∑ k = 0 n 1 k ! + 1 n ⋅ n ! {\displaystyle b_{n}=\sum _{k=0}^{n}{\tfrac {1}{k!}}+{\tfrac {1}{n\cdot n!}}} define nested intervals ( I n ) n ∈ N = ( [ a n , b n ] ) n ∈ N {\displaystyle (I_{n})_{n\in \mathbb {N} }=([a_{n},b_{n}])_{n\in \mathbb {N} }} , where the real number included in all intervals is exactly e {\displaystyle e} .
The advantage if the exponential series compared to the sequences defining e {\displaystyle e} is that one can achieve much faster convergence. For instance, with 10 elements ∑ k = 0 10 1 k ! ≈ 2.7182818011 {\displaystyle \sum _{k=0}^{10}{\tfrac {1}{k!}}\approx 2.7182818011} which is an approximation precise up to 7 digits: e = 2.718281828 {\displaystyle e=2.718281828} . By contrast, the 1000-th sequence element ( 1 + 1 1000 ) 1000 = 2.7169239 {\displaystyle \left(1+{\tfrac {1}{1000}}\right)^{1000}=2.7169239} is precise to only 2 digits after the comma. Outlook: generalized exponential series
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