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If $\{a_{n}\}>0$ and $\sum\limits_{n=1}^{\infty}a_n$ diverge.

The following series: converge, diverge, or neither?

$\sum\limits_{n=1}^{\infty} \dfrac{a_n}{1 + a_{n^2}} , \sum\limits_{n=1}^{\infty} \dfrac{a_n}{1 + na_{n}}$ and $\sum\limits_{n=1}^{\infty} \dfrac{a_n}{a_{n} +n^2 a-{n}}$ ?

1) $ \sum\limits_{n=1}^{\infty} \dfrac{a_n}{1 + na_{n}}$

let $ a_{n} = \frac{1}{n}.$ then $ \frac{a_n}{1 + na_{n}} = \frac{1}{2n} $

$ \sum\limits_{n=1}^{\infty} \dfrac{a_n}{1 + na_{n}} = $ $\sum\limits_{n=1}^{\infty} \frac{1}{2n}$ diverge .

Is this reasoning correct?

VERA
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    The wording of your question is unclear. I understood your question to mean, if $\sum a_n$ diverges, then does the sum $\sum a_n/(1 + a_{n^2})$ necessarily converge, necessarily diverge, or neither? – JavaMan Oct 19 '18 at 21:12
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    @JavaMan: it may be that the OP doesn’t understand the question correctly. I read it almost the same as you (with exception of applying the question to all three series). – Clayton Oct 19 '18 at 21:24
  • In the third one, there's a typo I think. ($a$ is undefined) –  Oct 19 '18 at 22:05
  • The above comments are right I think, and the question needs to be restated. – zhw. Oct 19 '18 at 22:08
  • Refer to the related OP – user Oct 19 '18 at 22:29

2 Answers2

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This reasoning is partially correct. By giving this "solution", you just proved that $\sum_{n=1}^{\infty} \frac{a_n}{1 + n a_n}$ does not necessarily converge when $\sum_{n=1}^{\infty} a_n$ diverges, but not that $\sum_{n=1}^{\infty} \frac{a_n}{1 + n a_n}$ necessarily diverges.

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I'll assume the question is as follows:

Let $\{a_n\}_{n=1}^\infty$ be a sequence satisfying the following properties: \begin{array}{c} (1)\quad a_n>0\text{ for all }n\\ (2)\quad \displaystyle\sum_{n=1}^\infty a_n\quad \text{diverges}. \end{array} Do each of the following series necessarily converge, necessarily diverge, or neither? $$(A)=\sum_{n=1}^\infty\frac{a_n}{1+a_{n^2}},\qquad(B)=\sum_{n=1}^\infty\frac{a_n}{1+na_n},\qquad(C)=\sum_{n=1}^\infty\frac{a_n}{a_n+n^2a_n}.$$

(Note the slight change in $(C)$ as it appears to be a typo in the original post.) For $(A)$, we can easily check demonstrate divergence by taking $a_n=\frac1n$ and likewise, we can demonstrate convergence by taking $a_n=e^n$. Clearly both conditions are satisfied for the choices of $a_n$, thus we've shown that $(A)$ doesn't necessarily converge or diverge; that is, convergence/divergence depends on $a_n$.

User gimusi posted in the comments that this answer by Did settles $(B)$ entirely - it neither necessarily converges nor necessarily diverges.

For $(C)$, we have $$\frac{a_n}{a_n+n^2a_n}\leq\frac{a_n}{n^2a_n}=\frac{1}{n^2}.$$ Thus, by direct comparison with the $p$-series $\sum_{n=1}^\infty n^{-2}$, we see that $(C)$ necessarily converges.

Clayton
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