sequences and series - Convergence of $sum_{n=2}^{infty}frac{(-1)^n+log(1+n^p)}{sqrt{n-sin n}}$

$\sum_{n=2}^{\infty}\frac{(-1)^n+\log(1+n^p)}{\sqrt{n-\sin n}}$

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.

Answer

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.
$$\frac{1+\log(1+(2n)^p)}{\sqrt{2n-\sin 2n}}=\frac{1}{\sqrt{2n-\sin 2n}}+\frac{\log(1+(2n)^p)}{\sqrt{2n-\sin 2n}}\geq\frac{1}{\sqrt{2n-\sin 2n}}\geq \frac{1}{\sqrt{2n+1}}\ .$$