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? 
 A: 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,
$$\sum_{n=1}^N \sin n = \frac{2\sin(N/2) \sin((N+1)/2)}{\sin(1/2)}$$
A: i). $b_n = a_n / n$
$n$ is always positive in this summation, so $b_n$ and $a_n$ have the same sign.
If $\sum_{n=1}^{\infty} a_n$ converges absolutely, then $0 \leq |a_n|/n \leq |a_n| \ \ \ \forall n\geq 1$ so $ \sum b_n$ also converges absolutely by the squeeze or sandwich theorem. 
If  $\sum_{n=1}^{infty} a_n$ converges conditionally, then (a) $b_n$ also has strictly alternating signs, (b) $\lim_{n \rightarrow \infty} a_n = 0$ implies $\lim_{n \rightarrow \infty} a_n/n = b_n = 0$, (c) $ \forall n, |a_{n+1}| < |a_n|$ implies $\forall n, |a_{n+1}|/n < |a_n|/n$ and $b_{n+1}=|a_{n+1}|/(n+1) < |a_{n+1}|/n < |a_n|/n = b_n$  so $\sum b_n$ also converges conditionally.
(ii) $b_n = a_n \sin (n)$
If $a_n \geq 0  \ \ \ \forall n \geq 1,$ OR if $\sum a_n $ converges absolutely, then since $-1 \leq \sin (n) \leq 1 \implies 0 \leq |\sin (n) | \leq 1$ 
Thus $0 \leq |b_n| = |\sin (n)| |a_n| \leq |a_n| $ so $\sum b_n$ converges absolutely by the squeeze or sandwich theorem..
BUT if $\sum a_n$ converges only conditionally, in this case we canot be sure about $ \sum b_n$
(iii) $b_n = n^{1/n} a_n$
$ n = e^{\ln n} $ so $ n^{1/n} = e^{\ln n \cdot 1/n} = e^{\frac{\ln n}{n}}$
$ \ln n < n \ \ \ \  \forall n > 0$ and by l'Hopital or other methods $ \lim_{n \rightarrow \infty} (\ln n) / n = 0$, so $ \lim_{n \rightarrow \infty} e^{\frac{\ln n}{ n}} = e^0 = 1$
Thus $\lim_{n \rightarrow \infty} b_n = a_n$ and therefore $\sum b_n$ converges by the limit comparison test.
Yes there is a general method. Use the squeeze or sandwich theorem and comparison test where possible and use limits and the limit comparison tests otherwise. It helps to know properties of trig functions and exponents and other common functions in the limit. Be very careful with signs. 
A: For (iii) consider the sequence to be $n^{-n}$
