# Convergence/divergence of these series

$\{a_n\}$ is a sequence of positive real numbers such that $\sum_{n=1}^\infty a_n$ is convergent. Which of the following series are convergent?

1. $$\sum_{n=1}^\infty \frac{a_n}{1 + a_n}$$
2. $$\sum_{n=1}^\infty \frac{a_n^\frac{1}{4}}{n^\frac{4}{5}}$$
3. $$\sum_{n=1}^\infty na_n\sin\frac{1}{n}$$

1) looks convergent to me by the comparison test: $\frac{a_n}{1 + a_n} < a_n$ and $\sum_{n=1}^\infty a_n$ converges.

I have no idea what to do for 2) and 3). I have tried to use Dirichlet's test, but couldn't figure out anything that worked.

Edit: I just realized 3) could be done with the limit form of the comparison test: $n \sin \frac{1}{n}$ has positive terms for large enough $n$, and $$\lim_{n \rightarrow \infty} \frac{na_n \sin \frac{1}{n}}{a_n} = 1$$ I still don't know how to do 2) though.

## 1 Answer

• For 2, you can either use Hölder's inequality (which is a generalization of Cauchy-Schwarz inequality) to obtain

$$\sum_{n=1}^{\infty} \frac{a_n^{1/4}}{n^{4/5}} \leq \left( \sum_{n=1}^{\infty} a_n \right)^{1/4} \left( \sum_{n=1}^{\infty} \frac{1}{n^{16/15}} \right)^{3/4} < \infty$$

or use Young's inequality (which is a generalization of AM-GM inequality) to obtain

$$\sum_{n=1}^{\infty} \frac{a_n^{1/4}}{n^{4/5}} \leq \sum_{n=1}^{\infty} \left( \frac{1}{4}\cdot a_n + \frac{3}{4} \cdot\frac{1}{n^{16/15}} \right) < \infty.$$

• For 3, notice that $n \sin(1/n) \to 1$ as $n\to \infty$ and hence $n \sin(1/n)$ is bounded. So if $M > 0$ is a bound of $n \sin(1/n)$ then

$$\sum_{n=1}^{\infty} \left| n a_n \sin(1/n) \right| \leq \sum_{n=1}^{\infty} M a_n < \infty.$$

In fact, if you are aware of the inequality $\left|\sin x\right| \leq \left|x\right|$ which holds for all $x \in \mathbb{R}$ then you directly obtain

$$\sum_{n=1}^{\infty} \left| n a_n \sin(1/n) \right| \leq \sum_{n=1}^{\infty} a_n < \infty.$$