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Let $a_n$ be a series of non-negative real numbers. Suppose $\sum a_n $ diverges.

Prove :

If $\lim(na_n)$ exists (in $\mathbb R$ or $\infty$), then $\sum \dfrac{a_n}{1+na_n}$ diverges.

Thanks for helping!

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  • $\begingroup$ What are your thoughts? Have you made any progress so far? $\endgroup$
    – abiessu
    Jun 25, 2016 at 13:56
  • $\begingroup$ Yeah , i proved that if sum an diverges than sum1/1+an diverges too , couldn't seem how it could be helpful @abiessu $\endgroup$
    – user335501
    Jun 25, 2016 at 13:57
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    $\begingroup$ that does not seem true if you take $a_n=n^2$ then clearly $\sum a_n=\infty$ but $\sum \frac{1}{1+n^2}$ is convergent by comparison $\endgroup$
    – Spotty
    Jun 25, 2016 at 13:59
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    $\begingroup$ If $\lim_{n\to\infty} n a_n$ is finite then there exists a number $x \in \mathbb R$ so that $x≥na_n$ $\forall n$, then $\frac1{1+na_n}≥\frac1{1+x}$ and since $a_n$ is non-negative: $$\sum_n^N \frac{a_n}{1+na_n}≥\frac1{1+x}\sum_n^N a_n$$ $\endgroup$
    – s.harp
    Jun 25, 2016 at 14:00
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    $\begingroup$ @Spotty I think the second sum should be $\sum \frac{a_n}{1+na_n}=\sum \frac{n^2}{1+n^3}$, which diverges. $\endgroup$
    – Noble Mushtak
    Jun 25, 2016 at 14:19

1 Answer 1

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  • If $\lim na_n=0$ then clearly $a_n\sim \dfrac{a_n}{1+na_n}$ and the two series $\sum a_n$ and $\sum \dfrac{a_n}{1+na_n}$ have the same nature. Thus $\sum \dfrac{a_n}{1+na_n}$ is divergent.
  • If $\lim na_n=\ell$ with $\ell>0$ or $\ell=+\infty$ then $\frac{a_n}{1+na_n}\sim\frac{k}{n}$ where $k=1$ if $\ell=+\infty$ and $k=\ell/(1+\ell)$ otherwise. But $\sum\frac{1}{n}$ is divergent, so $\sum \dfrac{a_n}{1+na_n}$ will also be divergent.
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