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The definition that I have for $e$ is $$e:=\lim_{n\to\infty }\left(1+\frac{1}{n}\right)^n.$$ How can we get that $$e=\sum_{k=0}^\infty \frac{1}{k!}\ \ ?$$

My try

By Newton formula $$\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{1}{n^k},$$ but I don't know how to prove that

$$\lim_{n\to\infty }\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{1}{n^k}=\sum_{k=0}^\infty \frac{1}{k!}.$$

idm
  • 11,824

2 Answers2

3

As you say, by the Binomial Theorem, $$ \left(1+\frac1n\right)^n =\sum_{k=0}^n\binom{n}{k}\frac1{n^k}\tag{1} $$ For each $k\le n$, by Bernoulli's Inequality, $$ \begin{align} \frac{\binom{n+1}{k}\frac1{(n+1)^k}}{\binom{n}{k}\frac1{n^k}} &=\frac{n+1}{n-k+1}\left(1-\frac1{n+1}\right)^k\\ &\ge\frac{n+1}{n-k+1}\left(1-\frac{k}{n+1}\right)\\ &=1\tag{2} \end{align} $$ and for each $k\gt n$, $\binom{n}{k}\frac1{n^k}=0$. Thus, for each $k$, $$ \binom{n+1}{k}\frac1{(n+1)^k}\ge\binom{n}{k}\frac1{n^k}\tag{3} $$ Furthermore, since $$ \binom{n}{k}\frac1{n^k}=\frac1{k!}\cdot\overbrace{\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdots\frac{n-k+1}{n}}^{\text{$k$ terms}}\tag{4} $$ we have, for $n\ge k$, $$ \frac1{k!}\left(\frac{n-k+1}{n}\right)^k\le\binom{n}{k}\frac1{n^k}\le\frac1{k!}\tag{5} $$ Thus, by the Squeeze Theorem, $$ \lim_{n\to\infty}\binom{n}{k}\frac1{n^k}=\frac1{k!}\tag{6} $$ So, by $(3)$ and $(6)$, each term of the sum in $(1)$ increases to $\frac1{k!}$. Therefore, by the Monotone Convergence Theorem, $$ \begin{align} \lim_{n\to\infty}\left(1+\frac1n\right)^n &=\lim_{n\to\infty}\sum_{k=0}^n\binom{n}{k}\frac1{n^k}\\ &=\sum_{k=0}^\infty\frac1{k!}\tag{7} \end{align} $$

robjohn
  • 345,667
1

It's easier to use Taylor series at a=0 to get the function for e: Taylor series is a series which is used to plot graphs,etc. and can be found using: ((x^n)f(a)[nth derivative of f(x) at x=a])/n! When this term is added for n going from 0 to infinity you get the Taylor series and here since a=0(special form of Taylor Series called Maclaurin Series) all the derivatives and f(a) are 1, thus each term in the Taylor series reduces to (x^n)/n! . When this term is added from n=0 to n=infinity, you get the familiar result for e

Hiten
  • 312