<s>[INST] Check if this series is convergent or not I've been going crazy for a long time in determining if this series converges or diverges. Most likely it converges.
$\displaystyle \sum_{n=1}^\infty \left(\frac{n}{2}\, \sin\frac{1}{n}\right)^\frac{n^2+1}{n+2}$
I am stuck in the necessary condition.
I tried to solve in this way :
$\displaystyle \frac{1}{2}\lim_{n \to \infty} \left(\frac{\sin\frac{1}{n}}{\frac{1}{n}}\right)^\frac{n^2+1}{n+2} = \frac{1}{2}\lim_{n \to \infty} (1)^\infty$
To solve this indeterminate form, I tried to use the exponential in this way
$\displaystyle  \frac{1}{2}\lim_{n \to \infty} \left(e^{\frac{n^2+1}{n+2}log \left(\frac{sin \frac{1}{n}}{\frac{1}{n}}\right)}\right)$
In this way I always get an indeterminate form
$\displaystyle \frac{1}{2}\lim_{n \to \infty} (e^{\infty *0})$
could you kindly give me support for the resolution of the character of this series?
 [/INST] Using the ratio test
$$a_n
=
\left(
\frac{n}{2}
\,
\sin\Bigl(\frac{1}{n}\Bigr)
\right)
^{\large{\frac{n^2+1}{n+2}}}\implies
\log(a_n)=\frac{n^2+1}{n+2}\log\left(
\frac{n}{2}
\,
\sin\Bigl(\frac{1}{n}\Bigr)
\right)$$
For large $n$, compose Taylor expansion
$$\log(a_n)=\frac{n^2+1}{n+2}\bigg[-\log (2)-\frac{1}{6 n^2}+O\left(\frac{1}{n^4}\right) \bigg]$$
$$\log(a_n)=n \log (2)+2 \log (2)+\frac{-\frac{1}{6}-5 \log (2)}{n}+\frac{\frac{1}{3}+10 \log (2)}{n^2}+O\left(\frac{1}{n^3}\right)$$ Replace $n$ by $n+1$ and continue with long division
$$\log(a_{n+1})-\log(a_n)=-\log (2)+\frac{\frac{1}{6}+5 \log (2)}{n^2}+O\left(\frac{1}{n^3}\right)$$
Now
$$\frac {a_{n+1}}{a_n}=e^{\log(a_{n+1})-\log(a_n)}=\frac{1}{2}+\frac{\frac{1}{6}+5 \log (2)}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
</s>