How do we know $\sum_{i=0}^n\frac{x^n}{n!} $ converges to $ e^x $ for all x?

Question

$$\sum_{i=0}^n\frac{x^n}{n!} $$ I know the sum converges for all x but how do we know it converges to the expect value $e^x$.

This sum was derived as the Taylor series of $e^x$ around $0$. How do we know works when we move from zero?

This is the easiest example i came up with, question can of course be generalized for other infinite series

Answer 1

You can show convergence via the ratio test. Some definitions say that $e^x :=$ its Taylor series. It is easy to see then via term by term differentiation that $(e^x)' = e^x$.

The first definition I learned started with $e^x$ being the function such that it is its own derivative. From there you can derive the entire Taylor Series.

Furthermore, you should check that $e^{x+y} = e^x*e^y$ in this definition.

Answer 2

Using your definition $$e^x:=\lim_{n\to \infty }\left(1+\frac{x}{n}\right)^n$$

here a proof (it's really not complicated, but I agree it's long). So feel free to ask if something is not clear.

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Using Binomial theorem, we have $$\left(1+\frac{x}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{x^k}{n^k}=\sum_{k=0}^n\frac{x^k}{k!}\prod_{i=0}^{k-1}\left(1-\frac{i}{n}\right).$$ The last equality come from the fact that \begin{align*} \frac{n!}{(n-k)!}&=n(n-1)(n-2)\cdot (n-k+1)\\ &=n^k\left(1-\frac{1}{n}\right) \left(1-\frac{2}{n}\right)...\left(1-\frac{k-1}{n}\right)\\ &=n^k\prod_{i=0}^{k-1}\left(1-\frac{i}{n}\right). \end{align*} Set $$S_n=\sum_{k=0}^n\frac{x^k}{k!}\prod_{i=0}^{k-1}\left(1-\frac{i}{n}\right).$$ Set $$T_n=\sum_{k=0}^n \frac{x^k}{k!}.$$ Fix $n\in\mathbb N^*$. Then, for all $k\leq n$, then $$\prod_{i=0}^{k-1}\left(1-\frac{i}{n}\right)\leq 1.$$ Therefore $S_n\leq T_n$ for all $n$, and thus $$\sum_{k=0}^\infty \frac{x^k}{k!}=\lim_{n\to \infty }T_n\geq \lim_{n\to \infty }S_n=e^x.$$ For the converse inequality, let $m\leq n$. Then, $$\sum_{k=0}^m \frac{x^k}{k!}\prod_{i=0}^{k-1}\left(1-\frac{k}{n}\right)\leq \sum_{k=0}^n\frac{x^k}{k!}\prod_{i=0}^{k-1} \left(1-\frac{k}{n}\right)=S_n.$$

Therefore, $$T_m=\lim_{n\to \infty }\sum_{k=0}^m \frac{x^k}{k!}\prod_{i=0}^{k-1}\left(1-\frac{k}{n}\right)\leq \lim_{n\to \infty }S_n=e^x.$$ Finally, we get $$\sum_{k=0}^\infty \frac{x^k}{k!}=\lim_{m\to \infty }T_m\leq e^x,$$ and thus $$\sum_{k=0}^\infty \frac{x^k}{k!}=e^x.$$

Answer 3

This is a proof that doesn't use Taylor series. We start from this property:

$$f(x)=f'(x) \Leftrightarrow f(x)=c_1e^x$$

Let's derive that sum:

$$\sum_{n=0}^{\infty} \frac{d}{dx} \frac{x^n}{n!}=\sum_{n=1}^{\infty} n\frac{x^{n-1}}{(n)!}=\sum_{n=1}^{\infty} \ \frac{x^{n-1}}{(n-1)!}$$

The index of the sum changed because the derivative of the constant term is clearly $0$. However we can see that this is the same sum (the index are just moved by 1). This proves that the sum is $c_1e^x$ and by checking for $x=0$ we can easily determine that $c_1=1$ which concludes the proof.

:)

Answer 4

Because, for each $n\in\mathbb N$,$$e^x=\left(\sum_{k=0}^n\frac{x^k}{k!}\right)+\frac{e^cx^{n+1}}{(n+1)!},$$for some $c$ between $0$ and $x$ (the mean-value form of the remainder of Taylor polynomials). Now, use the fact that $e^c<1$ if $x<0$ and that $e^c<e^x$ if $x>0$. Finally, use the fact that both limits$$\lim_{n\to\infty}\frac{x^{n+1}}{(n+1)!}\text{ and }\lim_{n\to\infty}\frac{e^xx^{n+1}}{(n+1)!}$$are equal to $0$, no matter the value of $x$.