10. $[8]$ Evaluate the infinite sum
$$
\sum_{n=0}^{\infty}\binom{2 n}{n} \frac{1}{5^{n}}
$$
Answer: $\sqrt{5}$ First Solution: Note that
$$
\begin{aligned}
\binom{2 n}{n} & =\frac{(2 n)!}{n!\cdot n!}=\frac{(2 n)(2 n-2)(2 n-4) \cdots(2)}{n!} \cdot \frac{(2 n-1)(2 n-3)(2 n-5) \cdots(1)}{n!} \\
& =2^{n} \cdot \frac{(-2)^{n}}{n!}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right) \cdots\left(-\frac{1}{2}-n+1\right) \\
& =(-4)^{n}\binom{-\frac{1}{2}}{n} .
\end{aligned}
$$

Then, by the binomial theorem, for any real $x$ with $|x|<\frac{1}{4}$, we have
$$
(1-4 x)^{-1 / 2}=\sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n}(-4 x)^{n}=\sum_{n=0}^{\infty}\binom{2 n}{n} x^{n} .
$$

Therefore,
$$
\sum_{n=0}^{\infty}\binom{2 n}{n}\left(\frac{1}{5}\right)^{n}=\frac{1}{\sqrt{1-\frac{4}{5}}}=\sqrt{5}
$$

Second Solution: Consider the generating function
$$
f(x)=\sum_{n=0}^{\infty}\binom{2 n}{n} x^{n}
$$

It has formal integral given by
$$
g(x)=I(f(x))=\sum_{n=0}^{\infty} \frac{1}{n+1}\binom{2 n}{n} x^{n+1}=\sum_{n=0}^{\infty} C_{n} x^{n+1}=x \sum_{n=0}^{\infty} C_{n} x^{n},
$$
where $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$ is the $n$th Catalan number. Let $h(x)=\sum_{n=0}^{\infty} C_{n} x^{n}$; it suffices to compute this generating function. Note that
$$
1+x h(x)^{2}=1+x \sum_{i, j \geq 0} C_{i} C_{j} x^{i+j}=1+x \sum_{k \geq 0}\left(\sum_{i=0}^{k} C_{i} C_{k-i}\right) x^{k}=1+\sum_{k \geq 0} C_{k+1} x^{k+1}=h(x),
$$
where we've used the recurrence relation for the Catalan numbers. We now solve for $h(x)$ with the quadratic equation to obtain
$$
h(x)=\frac{1 / x \pm \sqrt{1 / x^{2}-4 / x}}{2}=\frac{1 \pm \sqrt{1-4 x}}{2 x} .
$$

Note that we must choose the - sign in the $\pm$, since the + would lead to a leading term of $\frac{1}{x}$ for $h$ (by expanding $\sqrt{1-4 x}$ into a power series). Therefore, we see that
$$
f(x)=D(g(x))=D(x h(x))=D\left(\frac{1-\sqrt{1-4 x}}{2}\right)=\frac{1}{\sqrt{1-4 x}}
$$
and our answer is hence $f(1 / 5)=\sqrt{5}$.