Am I right in my conclusions about these series? I'm trying to decide if these series converge or diverge: 


*

*$$\sum_{n=1}^{\infty} (-1)^n \left(\frac{2n + 100 }{3n + 1 }\right)^n $$ Here $\lim_{n\to\infty} \left(\frac{2n + 100 }{3n + 1 }\right)^n \ne 0$, so can we conclude that the series diverges? 

*$$ \sum_{n=1}^{\infty} \log \left( n \sin \frac{1}{n} \right) $$ Can we compare this series with the divergent series $ \sum \log \sin(1/n)$ to conclude that it is divergent too? 
Am I reaching the correct conclusion in each case? 
 A: *

*We have
$$\sqrt[n]{\left|(-1)^n \left(\frac{2n + 100 }{3n + 1 }\right)^n\right|}=\left(\frac{2n + 100 }{3n + 1 }\right)\to\frac{2}{3}<1$$
so by the ratio test the series is convergent.

*By the Taylor series we have
$$\sin\left(\frac 1 n\right)=\frac 1 n +O\left(\frac{1}{n^3}\right)$$
so
$$\log\left(n\sin\left(\frac1 n\right)\right)=\log\left( 1  +O\left(\frac{1}{n^2}\right)\right)=O\left(\frac{1}{n^2}\right)$$
hence the series is convergent by limit comparison to the Riemann convergent series.
A: Hints:
For the first series, use to root test to prove the series converges.
$$\lim_{n\to \infty}\sqrt[\large n]{\left|(-1)^n \left(\frac{2n + 100}{3n + 1}\right)^n\right|}=\lim_{n\to \infty}\left(\frac{2n + 100 }{3n + 1 }\right) = \frac23 < 1$$
Hence, $$\sum_{n = 1}^\infty (-1)^n \left(\frac{2n + 100}{3n + 1}\right)^n\;\;\text{converges}$$

For the second series, use the comparison text to prove the series converges, too.
Clarification:
We know by the Taylor series expansion that
$$\sin\left(\frac 1 n\right)=\frac 1 n +O\left(\frac{1}{n^3}\right).$$
It follows, then, that $$\log\left(n\sin\left(\frac1 n\right)\right)=\log\left(n \left(\frac 1n + O\left(\frac 1{n^3}\right)\right)\right) = 1 + O \left(\frac 1{n^2}\right) = O\left(\frac 1{n^2}\right)$$
So the series is convergent by comparison to $\sum \dfrac 1{n^2}$
