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Suppose $\displaystyle\sum_{n=1}^{\infty}a_{n}$ diverges. Does $\displaystyle\sum_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}}$ diverge?

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To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. Also, many would consider your post rude because it is a command to others to solve your problem, not a request for help, so please consider rewriting it. – Zev Chonoles Jan 3 '13 at 5:46
Are $a_n$'s positive? – user17762 Jan 3 '13 at 5:46
@ Marvis Not necessarily – aliakbar Jan 3 '13 at 5:47
@aliakbar If you answer to Marvis is not necessarily so, then the answer to your question is not necessarily so. $a_n = 1$ diverges, but $a_{2k} = 1, a_{2k+1} = \frac {-1}{2(2k+1)+1}$ converges as it gives us $\frac {a_n}{1+na_n} = \frac {(-1)^n}{n+1}$. – Calvin Lin Jan 3 '13 at 5:59
up vote 15 down vote accepted

The series does not necessarily diverge. For example, let $a_n$ be the indicator function of the squares, that is the function which is $1$ when $n$ is a the square of an integer, and $0$ otherwise.** Then the series $\sum_{n=1}^\infty a_n$ diverges, since it fails the divergence test, yet $$\sum_{n=1}^\infty \frac{a_n}{1+na_n}$$ converges by comparison to $\sum_{n=1}^\infty \frac{1}{n^2}$.

Remark: This is problem $11$ $(d)$ from chapter $3$ of Rudin's "Principles of Mathematical Analysis Third Edition." I'll add that I think this is a great problem. The solution is strikingly simple, yet most solvers take a very long time to arrive at it, and erroneously try to prove divergence.

** If you require that $a_n>0$ let $a_n=2^{-n}$ on the non-squares to arrive at the same end result.

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Nice one. Should have thought of that! – Calvin Lin Jan 3 '13 at 6:01
Indeed, a very Rudinesque problem. – Brian M. Scott Jan 3 '13 at 6:21

Eric has proven that the divergence of $$ \sum_{n=1}^\infty a_n\tag{1} $$ does not imply the divergence of $$ \sum\limits_{n=1}^\infty\frac{a_n}{1+na_n}\tag{2} $$ even if $a_n\ge0$.

However, if $a_n$ decreases monotonically to $0$, then $(2)$ also diverges.

First, note that $\displaystyle\frac{a_n}{1+na_n}$ also decreases monotonically to $0$ since $$ \frac{a_{n+1}}{1+(n+1)a_{n+1}}=\frac1{1/a_{n+1}+(n+1)}\le\frac1{1/a_n+n}=\frac{a_n}{1+na_n}\tag{3} $$ Next, note that for $x\ge0$, $$ \frac{x}{1+x}\ge\frac12\min\left(x,1\right)\tag{4} $$ Setting $x=na_n$ and dividing by $n$, $(4)$ becomes $$ \frac{a_n}{1+na_n}\ge\frac12\min\left(a_n,\frac1n\right)\tag{5} $$ Now, there are two cases:

  1. $a_n\ge\frac1n$ infinitely often

  2. there is an $N$ so that for $n\ge N\Rightarrow a_n\lt\frac1n$.

In case 1, we can generate a sequence $n_k$ so that $n_{k+1}\ge2n_k$ and $a_{n_k}\ge\frac1{n_k}$. Then $$ \begin{align} \sum_{n=n_k+1}^{n_{k+1}}\frac{a_n}{1+na_n} &\ge(n_{k+1}-n_k)\frac{a_{n_{k+1}}}{1+n_{k+1}a_{n_{k+1}}}\\ &\ge(n_{k+1}-n_k)\frac1{2n_{k+1}}\\ &\ge\frac14\tag{6} \end{align} $$ Since we can find infinitely many such $k_n$, we have that $(2)$ diverges.

In case 2, for $n\ge N$, we have $\frac{a_n}{1+na_n}\ge\frac{a_n}{2}$. Therefore, since $(1)$ diverges, $(2)$ also diverges.

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