$\sum_{n=2}^{\infty}\frac{(-1)^n+\log(1+n^p)}{\sqrt{n-\sin n}}$
A) For which $p\in \mathbb{R}$ is the series convergent?
B) For which $p\in \mathbb{R}$ is the series divergent, and what is the sum?
C) Is the series absolutely convergent for any $p\in \mathbb{R}$?
I don't even know how to start, so far when I had a problem with parameter I could solve by using some tests. But in this case I can't see any tests I would be able to use directly.
A-B) For $p\leq0$ $$\sum_{n=2}^{\infty}\frac{(-1)^n+\log(1+n^p)}{\sqrt{n-\sin n}}=\sum_{n=2}^{\infty}\frac{(-1)^n}{\sqrt{n-\sin n}}+\sum_{n=2}^{\infty}\frac{\log(1+n^p)}{\sqrt{n-\sin n}}\ , $$ the first term converges by Leibniz criterion and the second term can be written as $$\sum_{n=2}^{\infty}\frac{\log(1+n^p)}{\sqrt{n-\sin n}}=\sum_{n=2}^{\infty}\frac{n^p+O(n^{2p})}{\sqrt{n-\sin n}}\ ,$$ which conveges for $p<-\frac{1}{2}$ and diverges for $-\frac{1}{2}\leq p\leq0$.
For $p>0$ the numerator is positive so the series has only positive terms: you obtain divergence by comparison criterion $$\frac{(-1)^n+\log(1+n^p)}{\sqrt{n-\sin n}}\geq \frac{\log(1+n^p)-1}{\sqrt{n+1}}\geq \frac{C}{\sqrt{n+1}}\ .$$ C) Absolute convergence never occurs since for any $p$ $$ \left|\frac{(-1)^n+\log(1+n^p)}{\sqrt{n-\sin n}}\right| $$ has a subsequence whose sum is divergent i.e. $$\frac{1+\log(1+(2n)^p)}{\sqrt{2n-\sin 2n}}=\frac{1}{\sqrt{2n-\sin 2n}}+\frac{\log(1+(2n)^p)}{\sqrt{2n-\sin 2n}}\geq\frac{1}{\sqrt{2n-\sin 2n}}\geq \frac{1}{\sqrt{2n+1}}\ .$$