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I am looking at the exponential power series:

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

It is $R=\displaystyle{\frac{1}{\lim_{n \to \infty} \sup \sqrt[n]{|a_n|}}}=\frac{1}{\lim_{n \to \infty} \sup \sqrt[n]{n!}}=+\infty$

So,the power series converges at $(-\infty,+\infty)$,so $\exists s(x): \mathbb{R} \to \mathbb{R} \text{ such that } \displaystyle{s(x)=\sum_{n=0}^{\infty} \frac{x^n}{n!}}, x \in \mathbb{R}$

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

According to a theorem: $$s'(x)=\sum_{n=1}^{\infty} n \frac{1}{n!}x^{n-1}=\sum_{n=0}^{\infty} \frac{x^n}{n!}, x \in \mathbb{R}$$

So, $s(x)=s'(x), x \in \mathbb{R}$.

In my notes,they continue,defining $f(x)=e^{-x} s(x),x \in \mathbb{R}$ and they conclude that $f(x)=c, \forall x \in \mathbb{R}$

Then,using the fact that $s(1)=e$,they conclude that $s(x)=e^x, \forall x \in \mathbb{R}$.

Instead of defining a function, could I do it like that: $$s'(x)=s(x) \Rightarrow \frac{s'(x)}{s(x)}=1 \Rightarrow \ln|s(x)|=x+c \Rightarrow s(x)=Ce^x\text{ ??}$$

Or isn't it possible,because we don't know if it is $s(x)=0$ ?

Also,is it known that $\sum_{n=1}^\infty \frac{1}{n!}=e$ ?

Michael Hardy
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evinda
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    Yes, $\sum\dfrac{1}{n!}$ is $e$. –  May 18 '14 at 16:53
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    Also, note that $s(x)=0$ is a solution, but in general the solution is $Ce^x$, and $e^x=\sum\dfrac{x^n}{n!}$. –  May 18 '14 at 16:54
  • So,you mean that I could also do this,in the way that I suggested?? And how can I check the case $s(x)=0$ ? – evinda May 18 '14 at 16:55
  • You may check the case $s(x)=0$ by directly substituting it into $s^\prime=s$. –  May 18 '14 at 16:56
  • You mean that $s(x)=0$ is a solution,because $s'=s$ ? – evinda May 18 '14 at 16:57
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    Yes. ${}{}{}{}{}$ –  May 18 '14 at 16:58
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    It is NOT true that $\sum_{n=1}^\infty 1/n!=e$. But it is true that $\sum_{n=0}^\infty 1/n!=e$. The lower bound of summation is $0$, not $1$. (So that first sum is $e-1$.) ${}\qquad{}$ – Michael Hardy May 18 '14 at 17:04
  • A ok!!! But, as we are looking for $s(1)$,it is equal to $\sum_{n=0}^{\infty} \frac{1}{n!}=1$,right? :) – evinda May 18 '14 at 17:07
  • That is correct. And in doing arithmetic, notice that every tail of that series is bounded above by a geometric series, so you get both upper and lower bounds on $e$ if you truncate it and do the arithmetic. – Michael Hardy May 18 '14 at 17:07
  • I would show convergence of this series by using a ratio test rather than a root test. – Michael Hardy May 18 '14 at 17:09
  • @SanathDevalapurkar Thank you also very much!!! Just to know..Have you already studied at university,with only $14$ years or have you learnt it by yourself? – evinda May 18 '14 at 17:11
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    @evinda By myself - I'm just an average middle-schooler! –  May 18 '14 at 17:12
  • @MichaelHardy how can I conclude that $\sum_{n=1}^{\infty} \frac{1}{n!}$ using the ratio test??Or don't you mean this?? – evinda May 18 '14 at 17:12
  • @SanathDevalapurkar That's great!!! :) – evinda May 18 '14 at 17:13
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    @evinda Thanks! In regard to your comment directed to Michael Hardy, how can you conclude that $\sum_{1}\dfrac{1}{n!}$ what? –  May 18 '14 at 17:14
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    I meant the power series $\sum_{n=0}^\infty x^n/n!$, and as I said, I would use the ratio test to show convergence. I didn't mean the ratio test would actually find the sum; just that it would show convergence. You seemed to be using a root test to do that. – Michael Hardy May 18 '14 at 17:14
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    @evinda Note that $s(1)=e$. –  May 18 '14 at 17:14
  • @MichaelHardy I used this formula to find the radius of convergence..not to find the convergence.. – evinda May 18 '14 at 17:16
  • @evinda You can find the radius of convergence iff the sequence converges (which it does). –  May 18 '14 at 17:18
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    If you apply the ratio test to the power series, you find that the limiting ratio is $0$ regardless of the value of $x$, so the series converges regardless of the value of $x$. That implies the radius of convergence is infinite. – Michael Hardy May 18 '14 at 17:18
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    Often if you apply the ratio test to a power series in $x$, the limiting ratio turns out to be $c|x|$ where $c$ is a constant (and "constant" means not depending on $x$), and if it's a power series in $(x-a)$ you may get $c|x-a|$. That tells you the series converges if $c|x|<1$, so if $-1/c<x<1/c$, and diverges if $c|x|>1$. That gives you the radius of convergence. The ratio test is simple for series like the one you're working with here because of the way in which factorials cancel. – Michael Hardy May 18 '14 at 17:24
  • Isn't an other formula for the radius of convergence like that:

    $$ R=\frac{1}{\lim_{n \to +\infty} \frac{a_{n+1}}{a_n}}$$

    ? In our case $\frac{a_{n+1}}{a_n}=\frac{1}{n+1} \to 0$,so the radius of convergence is equal to $+\infty$.

    Or am I wrong?

     – evinda May 18 '14 at 17:28
  • Isn't the radius of convergence also independent from $x$ ? – evinda May 18 '14 at 17:57
  • The question assumes that there is some definition of $e$ and $e^{x}$ (for all real $x$) available beforehand and then tries to show that $e^{x} = s(x)$. From the consideration of function $e^{-x}s(x)$ I believe that the definition used for $e^{x}$ is that $e^{x}$ is a function which satisfies $dy/dx = y$ and $y(0) = 1$. Also I don't understand why do we need to use the fact $s(1) = e$. It is much simpler to get $f(x) = e^{-x}s(x)= 1$ by putting $x = 0$ and not $x = 1$. – Paramanand Singh May 25 '14 at 06:00

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