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Prove that for every convergent serie $\sum^{\infty}_{n=1} a_n$ such that $a_n\ge 0$ for every $n\in \Bbb N$. Exist a sequence ({$b_n$}) such that $\lim_{n\to \infty} b_n=0$ and $\sum^{\infty}_{n=1} a_nb_n$ converge.

I have tried that $\sum^{\infty}_{n=1} a_n$ converge so we know that exist $N$ such that for all $m,n$ with $m>n>N$ we have |$\sum^{m}_n a_n$|<ϵ. But I can not figure out how to proceed.

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  • $\begingroup$ Are you sure you stated this correctly? As stated you could just let $b_n=0$. A more interesting question is to do the same with a sequence such that $b_n\to\infty$. $\endgroup$ Commented May 23, 2018 at 14:34
  • $\begingroup$ I double check it $\endgroup$
    – Roger F
    Commented May 23, 2018 at 14:42
  • $\begingroup$ So could I just take $b_n=0$ and I would have proved the problem? $\endgroup$
    – Roger F
    Commented May 23, 2018 at 14:47

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Take any sequence $(b_n)$ which has positive terms and tends to $0$, e.g. $b_n=\frac1n$. If $n$ is large enough,$0\le b_n<1$ so $$0\le a_nb_n<a_n,$$ and $\sum_{n\ge 1} a_nb_n$ converges by the comparison theorem.

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  • $\begingroup$ Could you provide me the comparison theorem? $\endgroup$
    – Roger F
    Commented May 23, 2018 at 14:48
  • $\begingroup$ If series have positive terms, $\sum_n a_n$ converges and $b_n\le a_n$ for all $n$ large enough, then $\sum_n b_n$ converges too. $\endgroup$
    – Bernard
    Commented May 23, 2018 at 15:00

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