We have a positive series $\displaystyle\sum^\infty_{n=1}a_n$. is the following series converge or diverge ?$$\displaystyle\sum^\infty_{n=1}\frac{a_n}{1+n^2a_n}$$

Suppose $\displaystyle\sum^\infty_{n=1}a_n$ does converge, so by the comparsion test the given series also converge.

Suppose $\displaystyle\sum^\infty_{n=1}a_n$ does not converge:

  • If $a_n$ is a bounded sequence with a bound $M$ then:

    $\forall n \ a_n\le M \Rightarrow \large\frac{a_n}{1+n^2a_n}>\frac{a_n}{1+M}\to\infty$

    So the given series diverge.

  • If $a_n$ isn't bounded, it has a subsequence that tends to infinity, so we have: $\displaystyle\frac{a_{n_k}}{1+{n_k}^2a_{n_k}}\longrightarrow^{k\to\infty}\infty$ so the given series will diverge.

(Couldn't find the tex for the limit with arrow)


2 Answers 2


Expanding on the other answer because it was not immediately obvious to me, if $a_n = 0$, you can discard that term because it's $0$. So assume wlog that $a_n \neq 0$.

Then $\dfrac{a_n}{1 + a_nn^2} = \dfrac{1}{\frac{1}{a_n} + n^2} < \dfrac{1}{n^2}$, which gives you convergence always.


Hint: $\dfrac{a_n}{1 + n^2a_n} \leq \dfrac{1}{n^2}$

  • $\begingroup$ Even if $a_n$ diverge ? $\endgroup$
    – GinKin
    Apr 23, 2014 at 8:24
  • $\begingroup$ @GinKin: i think so because this is true regardless whether a_n converges or diverges. $\endgroup$
    – DeepSea
    Apr 23, 2014 at 8:25
  • $\begingroup$ @GinKin If $a_n$ is $0$, then the bound is trivial. If $a_n$ is nonzero, then multiply top and bottom by $1/a_n$. $\endgroup$
    – user98602
    Apr 23, 2014 at 8:27

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