If $\sum\limits_{n=1}^{\infty}a_{n}$ diverges, does $\sum\limits_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}}$ diverge?

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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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

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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