Radius of convergence and convergence sets of $\sum\limits_n\frac{2n+1}{(n-1)^2}x^n$ and $\sum\limits_n(-1)^n(\sqrt{n+1}-\sqrt{n})x^n$ I want to find the radius of convergence of the following series and the set of $x\in \mathbb{R}$ in which the series converge. 


*

*$$\sum_{n=2}^{\infty}\frac{2n+1}{(n-1)^2}x^n$$ 

*$$\sum_{n=0}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n})x^n$$ 


To find the radius of convergence we have to compute the limit $\lim_{n\rightarrow \infty}\sqrt[n]{|a_n|}$. 
Do we do something elese at the first case where the sum starts from $2$ and not from $0$ ? 
I have done the following: 


*

*$$\lim_{n\rightarrow \infty}\sqrt[n]{|a_n|}=\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{2n+1}{(n-1)^2}x^n\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{2n+1}{(n-1)^2}\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{2n+1}{n^2-2n+1}\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{\frac{2}{n}+\frac{1}{n^2}}{1-\frac{2}{n^2}+\frac{1}{n^2}}\right|}$$

*$$\lim_{n\rightarrow \infty}\sqrt[n]{|a_n|}=\lim_{n\rightarrow \infty}\sqrt[n]{\left|(-1)^n(\sqrt{n+1}-\sqrt{n})x^n\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\sqrt{n+1}-\sqrt{n}\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{1}{\sqrt{n+1}+\sqrt{n}}\right|}=|x|\lim_{n\rightarrow \infty}\sqrt[n]{\left|\frac{1}{\sqrt{n+1}+\sqrt{n}}\right|}$$ 
$$$$ 
Is this correct so far? How could we continue? 
 A: You are on the right track.  If one wishes to apply the root test, then the radius of convergence for the first problem is given by
$$\begin{align}
\frac1R&=\lim_{n\to \infty}\sqrt[n]{\frac{2n+1}{(n-1)^2}}\\\\
&=\lim_{n\to \infty}e^{\frac1n \log\left(\frac{2n+1}{(n-1)^2}\right)}\\\\
&=\lim_{n\to \infty}e^{\frac1n (\log(2n+1)-2\log(n-1))}\\\\
&=1
\end{align}$$
Therefore, the radius of convergence, $R$ is $1$

The radius of convergence for the second problem is given by
$$\begin{align}
\frac1R&=\lim_{n\to \infty}\sqrt[n]{\left|(-1)^n(\sqrt{n+1}-\sqrt n)\right|}\\\\
&=\lim_{n\to \infty}e^{-\frac1n \log\left(\sqrt{n+1}+\sqrt{n}\right)}\\\\
&=1
\end{align}$$
Therefore, the radius of convergence, $R$ is $1$
A: We can calculate the radius of convergence by the formula too: $R=lim|\frac{a_{n}}{a_{n+1}}|$, where $a_{n}$ is the $n^{\text{th}}$ term of the given series.
In your first problem, $a_n=\frac{2n+1}{(n-1)^2}$ and $a_{n+1}=\frac{2(n+1)+1}{n^2}$. So $\frac{1}{R}=lim\frac{(2n+3)(n-1)^2}{(2n+1)n^2}=lim\frac{(1+\frac{3}{2n})(1-\frac{1}{n})}{(1+\frac{1}{2n})}=1$. Hence Radius of convergence is $1$.
Similarly, we can calculate the second problem. 
