<s>[INST] Prove series convergence The problem states:
a) If $\sum_1^\infty {a_n}$ converges and ${b_n}=n^\frac{1}{n}{a_n}$, then $ \sum_1^\infty{b_n}$ converges, and
b) If $\sum_1^\infty {a_n}$ converges and ${b_n}=\frac{{a_n}}{(1+|{a_n}|)}$, then $\sum_1^\infty {b_n}$ converges.
For a) We have that ${c_n}=n^\frac{1}{n}$ is a stictly decreasing sequence (for $n>e$) and that it's bounded from above by $e^\frac{1}{e}$, so $\sum_1^\infty {a_n}{c_n}=\sum_1^\infty {b_n}$ converges( Abel's test).
For b) We have that ${c_n}=\frac{1}{(1+|{a_n}|)}$ is a monotone decreasing sequence, and it's bounded from above by 1, so $\sum_1^\infty {a_n}{c_n}=\sum_1^\infty {b_n}$ converges.
That's it, I would like to know if my proof is right, any help will be appreciated.
 [/INST] In (a) it's correct with one detail: since $c_n$ is decreasing, you should check whether it is bounded from below.
Problem (b) is a bit funny since it isn't true if we don't assume anything else, e.g. that $\displaystyle \sum_{n=1}^{\infty} a_n$ converges absolutely. Here's a counterexample: $a_n$ is the sequence of all terms in the following sum:
$$\sum_{k=1}^{\infty} \sum_{n=1}^{k^2} \left[ -\frac{1}{k} + \frac{1}{2k} + \frac{1}{2k} \right] = -1 + \frac{1}{2} + \frac{1}{2} \underbrace{- \frac{1}{2} + \frac{1}{4} + \frac{1}{4}}_{\text{4 times}} \underbrace{ - \frac{1}{3} + \frac{1}{6} + \frac{1}{6} }_{\text{9 times}} - \ldots $$
It clearly converges to $0$. The corresponding sum of $\displaystyle \sum_{n=1}^{\infty} b_n$ is 
$$\sum_{k=1}^{\infty} \sum_{n=1}^{k^2} \left[ \frac{ -\frac{1}{k} }{1+\frac{1}{k}} + \frac{ \frac{1}{2k} }{1+\frac{1}{2k}} + \frac{\frac{1}{2k}}{1+\frac{1}{2k}} \right] = \sum_{k=1}^{\infty} \sum_{n=1}^{k^2} \frac{1}{k} \left[ \frac{1}{1+\frac{1}{2k}} - \frac{1}{1+\frac{1}{k}}\right] \\ = \sum_{k=1}^{\infty} \sum_{n=1}^{k^2} \frac{1}{2k^2} \cdot \frac{1}{\left( 1+\frac{1}{k} \right) \left( 1+\frac{1}{2k} \right)} = \sum_{k=1}^{\infty} \frac{1}{2} \cdot \frac{1}{\left( 1+\frac{1}{k} \right) \left( 1+\frac{1}{2k} \right)}$$ 
which clearly diverges. 
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