Study of $\sum_{n=1}^{\infty}n^\alpha(\sinh{\frac{1}{n}}-\log{(n+1)}+\log{n}+\cos{\frac{1}{n}}-1)$ For $\alpha\in \mathbb{R}$ I want to study the behaviour of
$$\sum_{n=1}^{\infty}n^\alpha(\sinh{\frac{1}{n}}-\log{(n+1)}+\log{n}+\cos{\frac{1}{n}}-1)$$
I have thought to apply the asymptotic test but in order to be able to apply this I have to check if this is a series of positive term, so: how can I prove this?
Now:
$1)\sinh\frac{1}{n}=\frac{1}{n}+\frac{1}{6n^3}+o(\frac{1}{n^4})$
$2) \cos{\frac{1}{n}}=1-\frac{1}{2n^2}+o(\frac{1}{n^3})$
$3)\log({\frac{n}{n+1}})=-\log(1+{\frac{1}{n}})=-\frac{1}{n}+\frac{1}{2n^2}-\frac{1}{3n^3}+o(\frac{1}{n^3})$
And so:
$$n^\alpha(\sinh{\frac{1}{n}}-\log{(n+1)}+\log{n}+\cos{\frac{1}{n}}-1)\\ =n^\alpha(\frac{1}{n}+\frac{1}{6n^3}+o(\frac{1}{n^4})-\frac{1}{n}+\frac{1}{2n^2}-\frac{1}{3n^3}+o(\frac{1}{n^3})-\frac{1}{2n^2}+o(\frac{1}{n^3}))\\ =n^\alpha(\frac{-1}{6n^3}+o(\frac{1}{n^3}))=-\frac{1}{6n^{3-\alpha}}+o(\frac{1}{n^{3-\alpha}})$$
So the series converges when $3-\alpha<1$ and diverges when $3-\alpha\geq 1$.
Am I right?
For the first question I can say as suggested in an answer that $n^\alpha(\sinh{\frac{1}{n}}-\log{(n+1)}+\log{n}+\cos{\frac{1}{n}}-1)=-\frac{1}{6n^{3-\alpha}}+o(\frac{1}{n^{3-\alpha}})<0$, so the general term of the series is eventually negative. Am I right?

We have $$
n^\alpha\left(\frac{1}{n^2+n}-\frac{2n+1}{2(n^2(n+1))}+o\left(\frac{1}{n^2}\right)\right)=\frac{n^\alpha}{n(n+1)}\left(-\frac1{2n}+o\left(\frac{1}{n^2}\right)\right)<0$$ for $n$ sufficiently large.    It's just as good to have all the terms negative as positive, of course, so you can proceed.
Perhaps I ought to add that I haven't checked your calculations in detail.
