Radius of convergence $\sum_{n=1}^{\infty}\frac{x^n}{n+\sqrt{n}}$ I've been struggling to solve the following exercise:
For $x\in\mathbb{R}$, find the radius of convergence of the series $\sum_{n=1}^{\infty}\frac{x^n}{n+\sqrt{n}}$.
My approach so far: Compute $\limsup_{n\to\infty}\sqrt[n]{\frac{1}{n+\sqrt{n}}} = \limsup_{n\to\infty}\dfrac{1}{\sqrt[n]{n+\sqrt{n}}}$ in order to find the radius of convergence, but that leaves me with a sequence whose limit I haven't been able to find so far.
Thank you very much in advance.
 A: You could compute the limit of $\frac{(n+1)+\sqrt{n+1}}{n+\sqrt{n}}=\frac{1+1/n+\sqrt{1/n+1/n^2}}{1+\sqrt{1/n}}$. This tends to $1$. Therefore, $\sqrt[n]{n+\sqrt{n}}$ also tends to $1$.
A: Let's calculate the limit:
$$\lim_{n \to \infty} \sqrt[n]{\frac{x^n}{n+\sqrt{n}}}=
\lim_{n \to \infty} \frac{x}{\sqrt[n]{n+\sqrt[2]n}}=x$$
so the series converges, when $|x| < 1$.
A: Method 1. For $n\in \Bbb Z^+$ we have $$1<(n+\sqrt n)^{1/n}\le (n+n)^{1/n}=2^{1/n}\cdot n^{1/n}. $$ For $2\le n\in \Bbb Z^+$ let $n^{1/n}=1+x_n.$ Then $x_n>0,$ so $$n=(1+x_n)^n=1+\binom {n}{1}x_n+\binom {n}{2}(x_n)^2+...>\binom {n}{2}(x_n)^2,$$ so $\sqrt {\frac {2}{n-1}}=\sqrt {\frac {n}{\binom {n}{2}}}>x_n.$ So $x_n\to 0$ as $n\to \infty.$
Method 2. Elementary, without the Cauchy-Hadamard Radius Formula: Let $a_n=\frac {x^n}{n+\sqrt n}$ and $b_n=x^n.$
If $|x|<1$ then the series $\sum_nb_n$ is absolutely convergent, and we have $|a_n|\le |b_n|$ so $\sum_n a_n$ is (absolutely) convergent. 
If $|x|=1+r$ with $r>0$ then $|a_n|=$ $\frac {(1+r)^n}{n+\sqrt n}\ge$ $ \frac {(1+r)^n}{2n}\ge$ $\frac {1+rn}{2n}>$ $\frac {rn}{2n}=\frac {r}{2}$ so the terms of the series $\sum_na_n$ do not tend to $0.$
