Show that $\sum_{n=0}^\infty \frac{x^n}{(n!)^2}$ is strictly increasing.

Question

The assignment is:

Let $$f: \mathbb{R} \rightarrow \mathbb{R}, x \rightarrow \sum_{n=0}^\infty \frac{x^n}{(n!)^2}$$ and show that, $f\mid_{[0,\infty)}$ is strictly increasing with $f([0,\infty)) = [1,\infty]$

What I've got: $$\sum_{n=0}^\infty \frac{x^n}{(n!)^2} = \sum_{n=0}^\infty \frac{x^n}{n!} \cdot \frac{1}{n!}$$ $$\sum_{n=0}^\infty \frac{x^n}{n!} = \exp(x) > 0\ and \sum_{n=0}^\infty \frac{1}{n!} > 0 \ ,\forall x \in \mathbb{R}$$ $f(0) = 1$, and it follows that:

$$\frac{x^n}{(n!)^2} > 0$$ and $$\sum_{n=0}^\infty \frac{x^n}{(n!)^2} > 1$$

Now I need to show that for $u, v \in \mathbb{R_+}$ with $u > v$ that $f(u) > f(v)$ but I have no clue how to rearrange $\sum_{n=0}^\infty \frac{u^n}{(n!)^2}$ to get the inequality I want.

Edit: A try, but I don't think this counts as an answer:

$$ \sum_{n=0}^\infty \frac{u^n}{(n!)^2} > \sum_{n=0}^\infty \frac{v^n}{(n!)^2} $$

$$ \leftrightarrow \sum_{n=0}^\infty \frac{u^n}{n!} \cdot \frac{1}{n!} > \sum_{n=0}^\infty \frac{x^n}{n!} \cdot \frac{1}{n}$$

$$ \leftrightarrow \sum_{n=0}^\infty \frac{u^n}{n!} > \sum_{n=0}^\infty \frac{v^n}{n!}$$

$$ \leftrightarrow \exp(u) > \exp(v)$$

Edit: We have not discussed derivatives yet but we have recently discussed the main characteristics of $\exp(x)$.

Answer 1

For the first part, notice that if $0\leqslant x\lt y$, then for each $n\geqslant 1$, $x^n\lt y^n$, then we divide by $n!^2$ on both sides.

For the second part, use and show the following facts:

• $f(0)=1$,
• $f$ is continuous on $[0,\infty)$,
• $f(x)\geqslant x$ for each $x$.

Answer 2

Notice that the function $x\mapsto x^n$ is increasing for all $n\in\mathbb N$ hence $f$ is also increasing function.

We have clearly $f(0)=1$ and since $f(x)\ge 1+x$ then $\lim_{x\to\infty}f(x)=+\infty$.