Which of the following is divergent? $\sum\frac1n\sin^2\frac1n$, $\sum\frac1{n^2}\sin\frac1n$, $\sum\frac1n\log n$, $\sum\frac1n\tan\frac1n$ 
Which of the following is divergent?
(a) $\displaystyle \quad\sum_{n=1}^{\infty}\frac{1}{n}\sin^2(\frac{1}{n})$
(b) $\displaystyle \quad\sum_{n=1}^{\infty}\frac{1}{n^2}\sin(\frac{1}{n})$
(c) $\displaystyle \quad\sum_{n=1}^{\infty}\frac{1}{n}\log n$
(d) $\displaystyle \quad\sum_{n=1}^{\infty}\frac{1}{n}\tan (\frac{1}{n})$

My attempt:
$\sin(x)\le x$ so for option (b) $\sum_{n=1}^{\infty}\frac{1}{n^2}\sin(\frac{1}{n})\le\sum_{n=1}^{\infty}\frac{1}{n^2}$ which is convergent. Hence, series in (b) is convergent.
For option (a)
$\sum_{n=1}^{\infty}\frac{1}{n}\sin^2(\frac{1}{n})\le\sum_{n=1}^{\infty}\frac{1}{n^3}$ . However, I am not sure if the inequality I have used is correct. Can someone please check this and explain how to workout for options (c) and (d).
Thanks for the help !
 A: You have$$\lim_{n\to\infty}\frac{\frac1n\sin^2\left(\frac1n\right)}{\frac1{n^3}}=1$$and therefore the first series converges.
You always have (unless $n$ is $1$ or $2$) $\frac{\log n}n>\frac1n$ and therefre the third series diverges.
And$$\lim_{n\to\infty}\frac{\frac1n\tan\left(\frac1n\right)}{\frac1{n^2}}=1.$$So, the fourth series converges.
A: As these are series with positive terms, the simplest uses asymptotic equivalents:
We know that $\sin \frac1 n,\tan\frac1n\sim_\infty\frac1n$, so
$$\frac1n\,\tan\frac1 n\sim_\infty\frac 1{n^2},$$
which is a convergent $p$-series (the same argument would work for $a)$ and $b)$ as well).
As to $d)$, just observe that
$$\frac{\log n}n>\frac1n\quad\text{ if $n\ge 3$}.$$
A: Use absolute values for your first inequality.
For (c) we use integral comparison :
$$ \sum_{n=1}^{\infty}\dfrac{\ln(n)}{n} \sim \int_{1}^{\infty}\dfrac{\ln(t)}{t}dt=+\infty$$
hence (c) is divergent.

For (d) use :
$$ \tan(\frac{1}{n})<\dfrac{1}{n} $$ and the positivity of $\tan$ above $0$.
So (d) is convergent.
