$\sum_{n=1}^{\infty} n \sin\left({1\over n}\right)$ converges Question_

Does $$\sum_{n=1}^{\infty} n \sin\left({1\over n}\right)$$ converge or diverge?

When I use the Taylor series:
$$\sin\left({1 \over x}\right)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k-1)!x^k}$$
$$x\sin\left({1 \over x}\right)=1-{1\over 3!x^2}+{1\over5!x^4}-\cdots$$
Although
$$\sum_{n=1}^{\infty}{1\over x^m}$$ converges  if $m>1$, $\sum{1}$ diverges, so it is impossible to explicit all the terms.
Then, how can I check the convergence(or divergence) of the series? Also, I've got stuck in a similar series:
 $$\sum_{n=1}^{\infty} n \tan\left({1\over n}\right)$$
Could you please help me to get in the right way? Thanks.
 A: Since$$\lim_{n\to\infty}n\sin\left(\frac1n\right)=\lim_{n\to\infty}\frac{\sin\left(\frac1n\right)}{\frac1n}=1,$$your series diverges.
A: Since $\sin{x} \geq \frac{x}{2}$ for $0 \leq x < \frac{\pi}{2}$:
$$
\sum_{n=1}^\infty n\sin\left(\frac{1}{n}\right) \geq \sum_{n=1}^\infty n\left(\frac{1}{2n}\right) = \sum_{n=1}^\infty \frac{1}{2} = \infty
$$
For $\sum_{n=1}^\infty n\tan\left(\frac{1}{n}\right)$, we have $\tan{x} \geq x$ for $0 \leq x \leq \frac{\pi}{2}$.
A: The comparisson test says:

if $\lim_{n\to\infty} \frac{a_n}{b_n} = c$, and c is finite and positive, both $\sum_{n = 0}^{\infty} a_n$ and $\sum_{n = 0}^{\infty} b_n$ diverge or both converge.

So, in our case, take $a_n = n \sin(1/n)$ and $b_n = n$, we conclude that since $\lim_{n \to\infty} a_n/b_n = \infty$ then we conclude that the $a_n$ series diverges since the series for $b_n$ diverges, and this is given because:

if $\lim_{n \to\infty} a_n/b_n = \infty,$ if $\sum_{n = 0}^{\infty} b_n$ diverges, then $\sum_{n = 0}^{\infty} a_n$ diverges.

And to complete it:

if $\lim_{n \to\infty} a_n/b_n = 0,$ if $\sum_{n = 0}^{\infty} b_n$ converges, then $\sum_{n = 0}^{\infty} a_n$ converges.

