Am I right in my conclusions about these series?

Question

I'm trying to decide if these series converge or diverge:

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

2. $$ \sum_{n=1}^{\infty} \log \left( n \sin \frac{1}{n} \right) $$ 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?

Answer 1

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 $\sum \dfrac 1{n^2}$

Answer 2

1. 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.

2. 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.