convergence of $\sum_{n=1}^{\infty}(n^2+1)\log^\alpha \left ( \frac{ \sinh \frac{1}{n}}{\sin\frac{1}{n}} \right )$ $ \sum_{n=1}^{\infty}(n^2+1)\log^\alpha \left ( \frac{\sinh\frac{1}{n}}{\sin\frac{1}{n}} \right ) $
Find the value of $\alpha$ for which the series converges
please, I have no idea to approach the solution
thanks.
 A: Hint. By using Taylor series expansions, as $n \to \infty$, we have
$$
\frac{\sinh \frac1n}{\sin \frac1n} = 1+\frac{1}{3 n^2}+O\left(\frac{1}{n^4}\right)
$$ and
$$
(n^2+1)\log^\alpha \left(\frac{\sinh\frac{1}{n}}{\sin\frac{1}{n}}\right)=\frac{1}{3^a n^{2\alpha-2}}+O\left(\frac{1}{n^{2\alpha}}\right)
$$ the given series is then convergent iff we have $2\alpha-2>1$.
A: Hint: use Taylor series, again, again, and again.
In detail:
From a Taylor series around $0$, since $\frac{1}{n}\xrightarrow[n\to\infty]{}0$:
$$\sinh \frac{1}{n} = \frac{1}{n} + \frac{1}{6n^3} + o\left(\frac{1}{n^3}\right) \tag{1}$$
and
$$\sin \frac{1}{n} = \frac{1}{n} - \frac{1}{6n^3} + o\left(\frac{1}{n^3}\right)\tag{2}$$
from which
$$
\frac{\sinh \frac{1}{n}}{\sin \frac{1}{n}}
= \frac{1+\frac{1}{6n^2}+ o\left(\frac{1}{n^2}\right)}{1-\frac{1}{6n^2}+ o\left(\frac{1}{n^2}\right)}
= 1+\frac{1}{3n^2}+ o\left(\frac{1}{n^2}\right).
$$
(Using $\frac{1}{1+x}= 1-x+o(x)$ around $0$). From the Taylor series $\log(1+x)=x+o(x)$ at $0$, we get
$$
\log^\alpha \frac{\sinh \frac{1}{n}}{\sin \frac{1}{n}}
= \log^\alpha\left( 1+\frac{1}{3n^2}+ o\left(\frac{1}{n^2}\right)\right)
= \left( \frac{1}{3n^2}+ o\left(\frac{1}{n^2}\right)\right)^\alpha
= \frac{1}{3^\alpha n^{2\alpha}}+ o\left(\frac{1}{n^2\alpha}\right)
$$
and finally
$$
(n^2+1)\log^\alpha \frac{\sinh \frac{1}{n}}{\sin \frac{1}{n}}
= \frac{1}{3^\alpha n^{2(\alpha-1)}}+ o\left(\frac{1}{n^{2(\alpha-1)}}\right).
$$
Theorems of comparison for positive series should then lead you to the conclusion: you need $2(\alpha-1)> 1$.
