Q:

Find the radius of convergence for the series $\sum_{n=1}^\infty c_nx^n$

Find the radius of convergence for the series
$\sum_{n=1}^\infty c_nx^n$,
where
$c_n= \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}$  
I know how to find the radius of convergence if $c_n$ was a specific function, but here it's not. I need to find the limit as $x \to \infty$ of $|\frac{c_{n+1}x^{n+1}}{c_nx^n}|$.
What am I supposed to do with this information $c_n= \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}$ ?

A:

$$\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}=\sqrt{n}\cdot\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{n+\frac{k}{n}}}\approx \sqrt{n}\int_{0}^{1}\frac{dx}{\sqrt{n+x}}=\frac{2\sqrt{n}}{\sqrt{n}+\sqrt{n+1}}\approx 1.$$
So we have that our coefficients approach one quite fast. That gives that the radius of convergence of the associated analytic function is one, too.

A:

Note
$$\tag 1 \frac{n}{\sqrt {n^2 + n}} \le c_n \le \frac{n}{\sqrt {n^2 + 1}}$$
for all $n.$ Since the right and left sides of $(1)$ $\to 1,$ $c_n \to 1,$ which implies $c_n^{1/n} \to 1.$ By the root test, the radius of convergence is $1.$

