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If $a_n>0$ and $\sum a_n$ diverges, what can be said about $\displaystyle \sum \frac{a_n}{1+na_n}$?

I cannot prove that it is convergent or divergent. I think it is convergent for some examples and also divergent for some. I proved that this is divergent when $\displaystyle a_n=\frac{1}{n}$. Can anyone give me an example for which the series is convergent?

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Hint: if $\sup na_n=M<+\infty$ then $\sum \frac{a_n}{1+na_n}\geq \sum a_n/(1+M)=+\infty$. If $na_n\to \infty$ then the denominator $1+na_n\leq 2na_n$, so $$\sum\frac{a_n}{1+na_n}\geq \sum \frac{a_n}{2na_n}...$$

Thus for a sequence $a_n$ such that $\sum a_n=+\infty$ but $\sum \frac{a_n}{1+na_n}<+\infty$, the limit $\lim na_n$ does not exist. A counterexample is: $$a_n=1 \text{ if } n \text{ is a power of } 2,\qquad a_n=1/2^n \text{ otherwise}. $$ Let $A:=\{n: n \text{ is power of }2\}$, $B:=\mathbb{N}\backslash A$. Clearly both sets are infinite, hence $\sum a_n=+\infty$. But $$\sum \frac{a_n}{1+na_n} = \sum_A \frac{a_n}{1+na_n}+\sum_B \frac{a_n}{1+na_n},$$ and $$\sum_B \frac{a_n}{1+na_n}=\sum_B \frac{2^{-n}}{1+n2^{-n}}<+\infty.$$ $$\sum_A \frac{a_n}{1+na_n} = \sum_{\{n \text{ is power of }2\}} \frac{1}{1+n}= \sum_{k=1}^{+\infty} \frac{1}{1+2^k}<+\infty. $$ Thus $\sum \frac{a_n}{1+na_n}$ converges.

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  • $\begingroup$ How do you say that $na_n$ has a supremum? $\endgroup$
    – user184652
    Oct 22, 2014 at 0:46
  • $\begingroup$ I do the 2 cases, $na_n$ bounded and $na_n\to+\infty$. Both give divergence. Note there is still something to complete, since these 2 cases are not exhaustive. $\endgroup$
    – Milly
    Oct 22, 2014 at 0:47
  • $\begingroup$ The answer is actually you can find convergency. Hint may have been misleading, sorry for that (actually it was to say "for counterexamples \lim na_n should not exist"). See edits. $\endgroup$
    – Milly
    Oct 22, 2014 at 1:25

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