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?
 A: Letting
$$
a_n
=
\left(
\frac{n}{2}
\cdot
\sin\Bigl(\frac{1}{n}\Bigr)
\right)
^{\large{\frac{n^2+1}{n+2}}}
$$
we have $a_n > 0$ for all positive integers $n$, so we get
$$
\lim_{n\to\infty}
|a_n|
^
\frac{1}{n}
=
\lim_{n\to\infty}
(a_n)
^
\frac{1}{n}
=
\lim_{n\to\infty}
\left(
\frac{1}{2}
\cdot
\frac
{\sin\Bigl({\large{\frac{1}{n}}}\Bigr)}
{\Bigl({\large{\frac{1}{n}}}\Bigr)}
\right)
^{{\large{\frac{n^2+1}{n^2+2n}}}}
=
\Bigl(
\frac{1}{2}
\cdot
1
\Bigr)
^1
=
\frac{1}{2}
$$
hence by the root test, ${\displaystyle{\sum_{n=1}^\infty a_n}}$ converges.
A: 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
$$\frac {a_{n+1}}{a_n}=e^{\log(a_{n+1})-\log(a_n)}=\frac{1}{2}+\frac{\frac{1}{6}+5 \log (2)}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
