Testing for the Convergence/Divergence of a series Given only that the series $\sum a_n$ converges, either prove that the series $\sum b_n$ converges or give a counterexample, when we define $b_n$ by,
i ) $\frac{a_n}{n}$
ii) $a_n \sin(n ) $
iii) $n^{\frac{1}{n}} a_n$
Is there any general approach to such questions? 

Counterexample (ii):
The series 
$$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty\frac{\sin n}{n},$$
converges by the Dirichlet test. Using $\cos 2n = 1 - 2 \sin^2 n$ we have 
$$\sum_{n=1}^m a_n \sin n = \sum_{n=1}^m\frac{\sin^2 n}{n} = \sum_{n=1}^m\frac{1}{2n}- \sum_{n=1}^m\frac{\cos 2n}{2n}$$
The first sum on the RHS diverges (harmonic series) and the second sum converges by the Dirichlet test. Hence, the sum on the LHS diverges.
I'll leave it to you to attempt part (i) and (iii) where a standard series test applies.  Hint -- the sequence  $1/n$ is monotone decreasing and $n^{1/n}$ is   eventually monotone decreasing.  Case (ii) stands apart because $\sin n$ is not monotone.
Supplement:  Deriving $\sum_{n=1}^N \sin n = \frac{\sin (N/2) \sin( (N+1)/2)}{\sin (1/2)}$
We will use the identities $\sin a = - \cos (\pi/2 + a)$ and $\sin(a+b) - \sin(a-b) = 2 \sin b  \cos a. $
We have
$$\begin{align}2 \sin(1/2) \sum_{n=1}^N \sin n &= -\sum_{n=1}^N [2 \sin(1/2) \cos(\pi/2 +n)] \\ &=  -\sum_{n=1}^N [ \sin(\pi/2 + n + 1/2) - \sin(\pi/2 +n - 1/2)] \end{align}.$$
Since the series on the RHS is telescoping, we get
$$\begin{align}2 \sin(1/2) \sum_{n=1}^N \sin n &= - [\sin(\pi/2 + N + 1/2) - \sin(\pi/2 + 1/2)] \\ &=  -[ \sin( N/2 + 1/2 + \pi/2 + N/2 ) - \sin(N/2 + 1/2 + \pi/2 - N/2)] \end{align}$$
Using $\sin(a+b) - \sin(a-b) = 2 \sin b  \cos a $ with $a = N/2 + 1/2 + \pi/2$ and $b = N/2$,
$$\begin{align}2 \sin(1/2) \sum_{n=1}^N \sin n &= - 2\sin(N/2) \cos(\pi/2 + (N+1)/2) \\ &= 2\sin(N/2) \sin((N+1)/2) \end{align}$$
Hence,
$$\sum_{n=1}^N \sin n = \frac{2\sin(N/2) \sin((N+1)/2)}{\sin(1/2)}$$
