Suppose that $\sum a_n $ is a divergent series of real numbers.

What can be said about the convergence of $$\sum \frac {a_n}{a_nn^p+1} \hspace {15pt}\text {?} $$

Here $p \in \mathbb {R} $ is unspecified, meaning that different values may lead to different solutions.

Note. This question (which comes from me, not a teacher) generalizes an exercise from Rudin. Though the exercise in Rudin (which I solved prior to asking this question) keeps $a_n $ non-negative, I would like to know what happens if this assumption is dropped, in addition to taking arbitrary $p $. I do appreciate the very good answers below for the non-negative case; however, this is why I have not yet accepted an answer.

I did not give my progress because I had not made any. I am asking a question about mathematical analysis, not requesting homework help.

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    $\begingroup$ What have you found out about $$\sum \frac{a_n}{a_n n^p + 1}$$ so far? $\endgroup$ – Clark Zinzow Jan 9 '15 at 7:23
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    $\begingroup$ It might be easier to think of the term in the following, equivalent form (if no $a_n$ is $0$) $$\frac{1}{n^p+\frac1{a_n}}$$ $\endgroup$ – Regret Jan 9 '15 at 7:31
  • $\begingroup$ Sorry, Mhenni Benghorbal, I wrote a comment which was false. If that was why you deleted your answer, please undelete it. $\endgroup$ – Regret Jan 9 '15 at 7:47
  • $\begingroup$ Uh, are all $a_n > 0$? Or do we have some nonpostive terms? $\endgroup$ – IAmNoOne Jan 9 '15 at 8:26
  • $\begingroup$ Please reopen this question. It should never have been closed. The context you claim is missing is clearly evident from the content of the question. Furthermore, this is a useful question (with a good, but still incomplete, set of answers) that will continue to be helpful to future users. $\endgroup$ – WQ Arielle Cosette Jan 9 '15 at 17:25

Note: We assume that $a_n > 0$, in keeping with the exercise in Rudin's "Principles of Mathematical Analysis" that seemed to inspire this question, namely Exercise 3.11(d). This question is a generalization of that exercise, in which $p = 1$ and $p = 2$ were the only two cases considered.

Note that, \begin{align*} \sum_{n=1}^\infty \frac{a_n}{a_n n^p + 1} & = \sum_{n=1}^\infty \frac{1}{n^p + \frac{1}{a_n}} \\ & \leq \sum_{n=1}^\infty \frac{1}{n^p}. \end{align*} We have that $\sum \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \leq 1$ by the following. If $p \leq 0$, then $\lim_{n \to \infty} \frac{1}{n^p} \neq 0$ for $p \leq 0$, hence $\sum \frac{1}{n^p}$ diverges. If $p > 0$, then we can use a theorem of Cauchy to get us to the series $$\sum_{k=0}^\infty 2^k \cdot \frac{1}{2^{kp}} = \sum_{k=0}^\infty 2^{(1-p)k}.$$ Given that $2^{1-p} < 1$ if and only if $1 - p < 0$, we have that $\sum \frac{1}{n^p}$ converges for $p > 1$ by comparison with the geometric series $$\sum_{n=0}^\infty (2^{1-p})^n = \frac{1}{1-2^{1-p}}.$$

Therefore, we have that $\sum \frac{a_n}{a_n n^p + 1}$ converges if $p > 1$.

Let $p \leq 1$. Suppose that we let $a_n = \frac{1}{n}$. Then $$\sum_{n=1}^\infty \frac{a_n}{n^p a_n + 1} = \sum_{n=1}^\infty \frac{\frac{1}{n}}{n^p \frac{1}{n} + 1} = \sum_{n=1}^\infty \frac{1}{n^p + n} \geq \sum_{n=1}^\infty \frac{1}{2n} = \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n},$$ which diverges.

We therefore have that $\sum \frac{a_n}{a_n n^p + 1}$ converges for any $a_n > 0$ if $p > 1$ and we can choose some $a_n > 0$ such that $\sum \frac{a_n}{a_n n^p + 1}$ diverges for all $p \leq 1$.

  • $\begingroup$ $a_n > 0$ is not obligatory. With $a_n = (-1)^n$, we have $\sum_n a_n$ diverging. I imagine that $a_n = sin(n)$ has the property $\endgroup$ – Arnaud Fouchet Jan 9 '15 at 8:24
  • $\begingroup$ If we do not have that $a_n > 0$, then this problem seems to get less interesting. I assumed that $a_n > 0$ since this problem is a generalization of a popular exercise in Rudin's "Principles of Mathematical Analysis", Exercise 3.11(d), where $a_n > 0$ is given. But if you want to turn the conversation to any such real, divergent series $\sum a_n$, I suppose that it could end up being interesting! Definitely not as clean though. $\endgroup$ – Clark Zinzow Jan 9 '15 at 8:28
  • $\begingroup$ I'll add a note to the beginning of my answer to let people know that I'm assuming that $a_n > 0$. $\endgroup$ – Clark Zinzow Jan 9 '15 at 8:29

Here is one observation if $a_n > 0 $, $\sum_{n} a_n $ diverges and $p>1$ then the series

$$ \sum \frac {a_n}{a_nn^p+1} $$


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    $\begingroup$ I'm sorry about that. $\endgroup$ – Regret Jan 9 '15 at 7:52
  • $\begingroup$ @Regret: It is ok. Thanks for the comment. $\endgroup$ – Mhenni Benghorbal Jan 9 '15 at 7:53

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