Alternating series example

Question

I have a problem with rather simple example:

$$\sum\limits_{n=0}^{\infty} (-1)^{n+1}\frac{10^n}{n!}$$

I have to tell if it's convergent. I know it is, but I don't know how to prove it. I was thinking of Leibniz's test, but this sequence doesn't decrease monotonically.

Thank you in advance.

Answer 1

We have that, by induction, for $n$ sufficiently large (base case $n=10$) and assuming (induction step)

$$\frac{10^{n+1}}{(n+1)!}<\frac{10^n}{n!}$$

we have

$$\frac{10^{n+2}}{(n+2)!}=\frac{10}{n+2}\frac{10^{n+1}}{(n+1)!}<\frac{10}{n+2}\frac{10^n}{n!}<\frac{10^{n+1}}{(n+1)!}$$

therefore $a_n\to 0$ is strictly decreasing and we can refer to alternating series test.

Answer 2

If we recall the identity $$e^x = \sum\limits_{n=0}^{\infty} \frac{x^n}{n!} $$

Then we can calculate (as the series above exists) $$-e^{-10} =-\sum\limits_{n=0}^{\infty}\frac{(-10)^n}{n!} = \sum\limits_{n=0}^{\infty} (-1)^{n+1}\frac{10^n}{n!} $$ So it is a finite value and the series in question converges.