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Convergence/divergence of $\sum\frac{a_n}{1+na_n}$ when $\sum a_n$ diverges.

Let $a_n$ be a non-negative sequence such that the series $\sum a_n$ does not converge. Could the series $\sum a_n/(1+na_n)$ be convergent?

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marked as duplicate by Martin Sleziak, t.b., Sasha, Jason DeVito, robjohn Aug 7 '12 at 19:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

How big are the terms for which $a_n < 1/n$? What about when $a_n \ge 1/n$? – mjqxxxx Aug 7 '12 at 18:11
up vote 1 down vote accepted

Hint: Try $a_n=1$ if $n$ is a power of $2$ and $0$ otherwise.

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Assume all indices are positive for $\{ a_n \}$.

If $n$ is one less than a power of two, let $a_n = 1$. Else, let $a_n = 0$. Then the series $\sum a_n$ does not converge since the $i$-th partial sum is the number of numbers that are one less than a power of two and are less than or equal to $i$, and there are infinitely many positive numbers that are one less than a power of two.

On the other hand, $\sum a_n/(1 + na_n) = 1/2 + 1/4 + 1/8 + \cdots = \sum 2^{-n} = 1$.

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If $\sum a_n = \infty$, then there exist $n_0: n\ge n_0 \Rightarrow a_n \ge 1$. Hence, \begin{equation} \infty = \sum_{n\ge n_0} \dfrac{a_n}{a_n(n+1)} \le \sum_{ n \ge n_0}\dfrac{a_n}{(na_n+1)}. \end{equation} Then $\sum a_n/(1 + na_n)$ does not converge.

I messed up. My argument is if $\lim a_n=\infty$.

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What if $a_n=\frac{1}{2}$ (or even $a_n=\frac{1}{n}$) for all $n$? – Steven Stadnicki Aug 7 '12 at 18:34

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