TITLE: $f(x)=\displaystyle \sum\limits_{n=0}^{\infty}(-1)^n\frac{x^n}{n!n!}$ decreasing?
QUESTION [7 upvotes]: Define a function $f: \Bbb{R} \to \Bbb{R}$ by  $f(x)=\displaystyle \sum\limits_{n=0}^{\infty}(-1)^n\frac{x^n}{n!n!}$
Show that f is decreasing on on  $x \in [0,2]$ and that there exists a unique $x_0 \in [0,2]$ for which $f(x_0)=0$.
I started by calculating $f'$ but I do not know what to do next.

REPLY [4 votes]: The derivative is of the form $\sum (-1)^n a_n(x)$, where you can prove that $0 \leq a_{n+1}(x) \leq a_n(x)$ for every $n \geq 1$ and $x \in [0,2]$. Such an alternating series has the same sign as its first term (prove it!).

In order to prove that $f(2) < 0$, you can start noticing that $f(2) = 1 - 2 + 1 + R$ with
$$
R = \sum_{n=3}^\infty (-1)^n\frac{2^n}{(n!)^2}
$$
and justify why $R < 0$.