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Let $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ be two real sequences such that

  • $a_n\geq0$ and $b_n\geq0$ for all $n\in\mathbb N$;
  • $\sum_{n=1}^{\infty} a_n=\infty$;
  • $(b_n)_{n\in\mathbb N}$ is non-increasing ($b_1\geq b_2\geq\ldots$);
  • $\lim_{n\to\infty} b_n=0$; and
  • $\sum_{n=1}^{\infty}a_nb_n<\infty$.

I want to show that $\sum_{n=1}^Na_n=o(b_N^{-1})$, that is, $$\lim_{N\to\infty}\left\{b_N\sum_{n=1}^N a_n\right\}=0.$$ In words, the sequence $(b_n)_{n\in\mathbb N}$ converges so quickly that it not only soothes but suppresses the divergence of $\sum_{n=1}^Na_n$.

Any hints would be much appreciated.


My work: it is fairly clear that the sequence in question is bounded, that is, $\sum_{n=1}^N a_n=O(b_N^{-1})$, as for any $N\in\mathbb N$, one has that $$b_N\sum_{n=1}^N a_n=\sum_{n=1}^N a_nb_N\leq\sum_{n=1}^Na_n b_n\leq\sum_{n=1}^{\infty}a_nb_n(<\infty),$$ but I can’t seem to prove convergence to $0$.

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    $\begingroup$ This is reminding me of summation by parts ... $\endgroup$ – Michael Burr May 1 '16 at 22:39
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I got it. Any feedback on my solution would be appreciated.


Fix $\varepsilon>0$. Since $\sum_{n=1}^{\infty}a_n b_n<\infty$, there exists some $N_0\in\mathbb N$ such that $$\sum_{n=N_0+1}^{\infty}a_n b_n<\frac{\varepsilon}{2}.$$ Since $\lim_{n\to\infty} b_n=0$, there is some $N_1\in\mathbb N$ such that if $N\in\mathbb N$ and $N\geq N_1$, then $$b_N<\frac{\varepsilon}{2\left(\sum_{n=1}^{N_0} a_n+1\right)}$$ (the $+1$ added to deal with a potentially vanishing denominator). Hence, if $N\in\mathbb N$ and $N\geq \max\{N_0+1,N_1\}$, then one has that $$0\leq b_N\sum_{n=1}^Na_n= b_N\sum_{n=1}^{N_0}a_n+\sum_{n=N_0+1}^{N}a_n b_N<\frac{\varepsilon}{2}+\sum_{n=N_0+1}^{N}a_n b_n<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$ (Note the proof goes through the same way even if $\sum_{n=1}^{\infty} a_n<\infty$.)

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    $\begingroup$ I think you got it; looks perfect. $\endgroup$ – Fimpellizieri May 1 '16 at 23:48
  • $\begingroup$ @Fimpellizieri The solution looked too smooth as compared to the time it took me to figure it out, so I was a little skeptical about its validity—that’s why I appreciate the feedback. Thank you very much. $\endgroup$ – triple_sec May 1 '16 at 23:59

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