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.

  • $\begingroup$ In your case b), $c_n$ is not monotone decreasing... $\endgroup$ – Martigan May 13 '15 at 8:19
  • $\begingroup$ I meant that ${b_n}\ge{b_{n+1}}$ How is that called in english(not my first language) $\endgroup$ – mobzopi May 13 '15 at 8:24

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.

  • $\begingroup$ Ohh. You did it. Nice +1 $\endgroup$ – Bumblebee May 13 '15 at 9:11
  • $\begingroup$ Nice!, but why did Abel test fail here? $\endgroup$ – mobzopi May 13 '15 at 10:37
  • $\begingroup$ Because $c_n$ isn't decreasing. $\endgroup$ – Adayah May 13 '15 at 13:56
  • $\begingroup$ but ${c_n}$ just needs to be a monotone sequence $\endgroup$ – mobzopi May 13 '15 at 15:08
  • $\begingroup$ It isn't increasing too, so it's not monotone. $\endgroup$ – Adayah May 13 '15 at 15:35

For a) by limit comparison $\sum b_n$ is absolutely convergent$$\left|\dfrac{b_n}{a_n}\right|=n^{\frac1n}\to1$$ For b) $$a_n\to 0\implies |a_n|\lt 1\,\,\,\,\text{for sufficiently large}\,\,\,n\implies|b_n|\le\dfrac{|a_n|}{2}$$

  • 2
    $\begingroup$ What if $\displaystyle \sum_{n=1}^{\infty} a_n$ doesn't converge absolutely? $\endgroup$ – Adayah May 13 '15 at 8:26
  • $\begingroup$ Thank you. I have to fixed it. Again thank you. $\endgroup$ – Bumblebee May 13 '15 at 8:28

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