Proof of limit of sum of series

Question

I know that

$$\exp(-x)=\sum_{n=0}^\infty\frac{(-1)^{n+2}x^n}{n!}$$

converges for every positive $x$

Then we should have

$$\lim_{x\rightarrow+\infty}\sum_{n=0}^\infty\frac{(-1)^{n+2}x^n}{n!}=0$$

How can we prove the above result WITHOUT using the exponential function? I encoutered some similar situations where I have a series and I must calculate the limit of the sum. One of them is:

$$\lim_{x\rightarrow+\infty}\sum_{n=1}^\infty\frac{(-1)^{n+2}x^{n+1}}{n\cdot n!}$$

It's very helpful to solve one of them or both of them or all of them.

$(-1)^{n+2} = (-1)^n$; just saying. – taninamdar Sep 17 '14 at 16:56 — taninamdar, Sep 17 '14 at 16:56
I want to avoid $(-1)^0$ :-) – anonymous67 Sep 17 '14 at 17:02 — anonymous67, Sep 17 '14 at 17:02
And why would you want to do that? – taninamdar Sep 17 '14 at 17:07 — taninamdar, Sep 17 '14 at 17:07

score 2 · Answer 1 · answered Sep 17 '14 at 17:21

This prof is based on differential equations.

Let $f(x)=\sum_{n=0}^\infty(-1)^nx^n/n!$. Then $$ f'+f=0,\quad f(0)=1. $$ The equation $y'+y=0$ satisfies the conditions of existence and uniqueness of solution. Since $y\equiv0$ is a solution, uniqueness implies that $f(x)>0$ for all $x>0$. Then $f'<0$, $f$ is decreasing on $[0,\infty)$ and $\lim_{x\to\infty}f(x)=a\ge0$ exists. Suposse $a>0$. Then we would have $f'(x)=-f(x)<-a$ for all $x>0$. The would imply that $$ f(x)=1+\int_0^xf'(s)\,ds\le1-a\,x, $$ in contradiction with the fact that $f(x)>0$ for all $x>0$.

Proof of limit of sum of series

1 Answers1