Radius of convergence of $\sum_{n=1}^{\infty} \sin(\sqrt{n+1} - \sqrt{n})(x-2)^n$ $\sum_{n=1}^{\infty} \sin(\sqrt{n+1} - \sqrt{n})(x-2)^n$
There are two ways to determine the radius of convergence of the series:
$1. \lim_{n \rightarrow \infty}|\frac{a_n}{a_{n+1}}| $ and
$2. \lim_{n \rightarrow \infty}\sqrt[n]{a_n}$
But none of these work out easily, when i tried first one i ended up with $\frac{0}{0}$ and after using L'Hospitals rule it happened again so i couldn't find that limit, and i have no idea what could i do with the second one.
 A: $$
\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\sim\frac1{2\sqrt n}\quad\text{as }n\to\infty.
$$
Then
$$
\sin\bigl(\sqrt{n+1}-\sqrt{n}\bigr)\sim\sin\frac1{2\sqrt n}\sim\frac1{2\sqrt n}\quad\text{as }n\to\infty.
$$
A: Here's a different way to tackle the problem.
At $x=-1$ the power series trivially diverges so $r\leq 1$.
At $x=1$, $$\displaystyle \sin(\sqrt{n+1} - \sqrt{n})(-1)^n=\frac{(-1)^n}{2\sqrt{n}}-\frac 18 \frac{(-1)^n}{n^{3/2}}+o\left( \frac{1}{n^{3/2}}\right)$$
Let $\displaystyle u_n=\frac{(-1)^n}{2\sqrt{n}}$ and $\displaystyle v_n=-\frac 18 \frac{(-1)^n}{n^{3/2}}+o\left( \frac{1}{n^{3/2}}\right)$.
$\sum u_n$ converges via Leibniz test and $\sum v_n$ is absolutely convergent, hence convergent (some $\epsilon-N$ work is needed to prove absolute convergence).
Therefore, the power series converges at $x=1$, hence $r\geq 1$.
Finally $r=1$.
A: Let $A_n=\sqrt {n+1}\;-\sqrt n$ and $B_n=\sin A_n.$    We have $A_n=\frac {1}{2\sqrt n}(1+d_n)$ where $\lim_{n\to \infty}d_n=0.$ Since $\lim_{x\to 0}\frac {\sin x}{x}=1$ we have $B_n=\frac {1}{2\sqrt n }(1+e_n)$ where $\lim_{n\to \infty}e_n=0.$
So $1=\lim_{n\to \infty}|B_n|^{1/n}=$ $\lim_{m\to \infty}\sup_{n\geq m}|B_n|^{1/n}.$
The Cauchy-Hadamard Radius Formula: Let $(B_n)_{n\in \Bbb N}$ be any sequence in $\Bbb C.$ Let $S=\lim_{m\to \infty}\sup_{n\geq m }|B_n|^{1/n}.$  Let $R=1/S$ with the convention that if $S=0$ then $R=\infty,$ and if $S=\infty$ then $R=0.$
Then for any $z\in \Bbb C$ the series $\sum_{n=1}^{\infty}B_nz^n$ converges if $|z|<R$ and diverges if $|z|>R.$
Proof for $0<R<\infty:$
(i). Let $|z|=(1-k)R$ with $0<k\leq 1.$ For all but finitely many $n$ we have $|B_n|^{1/n} \leq(1+k)S=(1+k)\frac {1}{R}.$ This implies that $|B_nz^n|\leq ((1-k)(1+k))^n$ for all but finitely many $n,$ and we have $0\leq (1-k)(1+k)<1.$ 
(ii). Let $|z|=(1+k)R$ with $k>0.$ Take $k^* \in (0,1)$ such that $(1+k)(1-k^*)>1.$ There are infinitely many $n$ for which $|B_n|^{1/n}>(1-k^*)S=(1-k^*)\frac {1}{R}$.   So there are infinitely many $n$ for which $|B_nz^n|>((1+k)(1-k^*))^n,$ and we have $(1+k)(1-k^*)>1.$
In (ii), observe that if $|z|>R$ then  the terms of the series $\sum_n B_nz^n$ do not even converge  to $0$.
