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$$\sum_{i=0}^n\frac{x^n}{n!} $$ 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

Milan
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    What is your definition of $e^x$ (the answer really depend on that). – Surb Apr 20 '19 at 18:13
  • @Surb For the sake of not letting you use an easy argument i say $ lim (1+1/n)^n$ – Milan Apr 20 '19 at 18:15
  • @Jam It is a duplicate. i could not find a question similar to mine . Searching for question is hard here – Milan Apr 20 '19 at 18:25
  • @MilanStojanovic It's often a bit easier to Google than to use the in-site search engine. It helps to think of the key terms you want (which can be tricky when you want an equation). So "Taylor series", "exponential" and "convergence" work here. Also the "possible duplicate" comment gets posted when users flag questions as duplicates - I didn't actually write that :) – Jam Apr 20 '19 at 18:27
  • i do google but nothing – Milan Apr 20 '19 at 18:28
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    I think you want the definition $e^x=\lim\limits_{n\to \infty} \left(1+\dfrac{x}{n}\right)^n$ where $x\in\mathbb{R}, n\in\mathbb {N} $. More generally the definition holds for $x\in\mathbb {C} $ and one can prove that it equals the series in question. The key here is that both the series and the limit satisfy the functional equation $f(x+y) =f(x) f(y) $. – Paramanand Singh Apr 22 '19 at 06:08

4 Answers4

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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.

fGDu94
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  • Thats ok, but i would have to know that $e^x $ is the only function that satisfies those 2 properties ( a proof which i dont know).I would like to get a more general answer for checking convergence of maclaurin series to the original fucntion for all x – Milan Apr 20 '19 at 18:20
  • How will you define your $e^x$? – fGDu94 Apr 20 '19 at 19:48
  • no, i was just saying that idk if $e^x$ is the only function that satisfies that. currently with my limited knowledge i like the limit definition – Milan Apr 20 '19 at 21:20
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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.

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 $$\sum_{k=0}^\infty \frac{x^k}{k!}=e^x.$$

Surb
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  • Great answer. I would like to learn a general method to check if Maclaurin series converges to original function for all x. i used $e^x$ for simplicity ( that was a mistake). – Milan Apr 20 '19 at 18:22
  • Unfortunately you reverse a bit the process... Given a convergent series $\sum_{i}a_i(x)x^i$, it's normally not possible to give a closed form. Conversely, given an analytic function, it's very difficult (even impossible) to give a closed form of the entire series that converges to the function... Just for an example : what would be the series that converges to $\tan(x)$ ? (We know this one, but is very far than to be obvious). There are some exception as $\arctan$, $\sin$, $\cos$, $\exp$... but these are much more exception than normal fact. – Surb Apr 20 '19 at 18:29
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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.

:)

Kandinskij
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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 $x$.

José Carlos Santos
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