Question is :

For a sequence $\{a_n\}$ of positive terms, Pick out the cases which imply convergence of $\sum_{n=1}^{\infty} a_n$.

  • $\lim_{n\rightarrow \infty} n^{\frac{3}{2}}a_n=\frac{3}{2}$
  • $\sum_{n=1}^{\infty} n^2a_n^2<\infty$
  • $\dfrac{a_{n+1}}{a_n}< (\frac{n}{n+1})^2$

For the first case $\lim_{n\rightarrow \infty} n^{\frac{3}{2}}a_n=\frac{3}{2}$ :

Suppose $a_n > \frac{3}{2}. (\frac{1}{n})^{\frac{3}{2}}$ for all $n\in \mathbb{N}$ then we would have :

$n^{\frac{3}{2}}a_n > \frac{3}{2}.n^{\frac{3}{2}} (\frac{1}{n})^{\frac{3}{2}}=\frac{3}{2}$ but then $n^{\frac{3}{2}}a_n$ would not converge to $\frac{3}{2}$

So, for large $n$ we should have $$a_n \leq\frac{3}{2}. \frac{1}{n^{\frac{3}{2}}}\Rightarrow \sum_{n=1}^{\infty}a_n \leq \sum_{n=1}^{\infty} \frac{3}{2}. \frac{1}{n^{\frac{3}{2}}}=\frac{3}{2}\sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}$$

Right hand side converges and by comparison test $\sum_{n=1}^{\infty}a_n$ should converge.

For the second case $\sum_{n=1}^{\infty} n^2a_n^2<\infty$ :

Suppose that $na_n> \dfrac{1}{\sqrt{n}}$ for all $n$ then we would have $$n^2a_n^2> \dfrac{1}{n}$$ but then it is given that $\sum_{n=1}^{\infty} n^2a_n^2$ which would imply that $\sum_{n=1}^{\infty}\dfrac{1}{n}$ converges which is a contradiction.

Thus we should have $na_n\leq \dfrac{1}{\sqrt{n}}$ i.e., $a_n\leq \frac{1}{n\sqrt{n}}=\frac{1}{n^{\frac{3}{2}}}$ for all $n$ large.

So, We have $$\sum_{n=1}^{\infty}a_n\leq\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}$$

right hand side is convergent so by comparison test $\sum_{n=1}^{\infty}a_n$ converges.

For third case $\dfrac{a_{n+1}}{a_n}< (\frac{n}{n+1})^2$ :

we would have $\lim_{n\rightarrow \infty} \dfrac{a_{n+1}}{a_n} <\lim_{n\rightarrow \infty} (\frac{n}{n+1})^2=1$

Thus, by ration test, $\sum_{n=1}^{\infty} a_n$ Converges absolutely and so is convergent.

My choice of bounds :

  • $a_n > \frac{3}{2}. (\frac{1}{n})^{\frac{3}{2}}$ for all $n$ in first case
  • $na_n> \dfrac{1}{\sqrt{n}}$ for all $n$ in second case

All these came to me just by choice and i would like to know if there is some sense behind this choice.

Please help me to clear this and make this solution a bit more clear.

Please do not give an alternative solution until this problem is fully verified.

Thank you.

  • 1
    $\begingroup$ Your third case is related to the ratio test. $\endgroup$
    – John
    Jan 16, 2014 at 10:43
  • $\begingroup$ Why do you add additional conditions on $\{a_n\}$ to get convergent series? You are supposed to investigate which of the 3 given conditions is sufficient for $\sum_{n=1}^\infty a_n$ to converge, aren't you? $\endgroup$
    – Christoph
    Jan 16, 2014 at 10:44
  • 1
    $\begingroup$ @John : Ah!!!! A very Biggg +1.... I like this :) $\endgroup$
    – user87543
    Jan 16, 2014 at 10:44
  • $\begingroup$ @ChristophPegel : I am not adding any additional conditions... I am deducing conditions on $a_n$ with given conditions.. $\endgroup$
    – user87543
    Jan 16, 2014 at 10:45
  • 1
    $\begingroup$ @TattwamasiAmrutam : Yes yes.. I would see that and solution for $c$ is actually trivial i had made it so complicated... I would edit that in the question.. I have edited that $c$ part... have a look at that if you are interested $\endgroup$
    – user87543
    Jan 23, 2014 at 11:19

2 Answers 2


For (a) , if we take $$\{b_n\}=\frac{1}{n^{\frac{3}{2}}}$$, then $$\lim_{ n\to \infty}\frac{a_n}{b_n}=lim_{n\to \infty}n^{\frac{3}{2}}a_n=\frac{3}{2}$$ which is finite . Hence by limit comparison test $\{a_n\}$ converges.

For (b) since $$\sum({na_n})(\frac{1}{n})\le \{\sum n^2a_n^2\}^{\frac{1}{2}}\{ \sum\frac{1}{n^2}\}^{\frac{1}{2}}$$.

Hence $\{a_n\}$ is convergent.

  • 1
    $\begingroup$ your intention and idea are absolutely fine.. I love your idea for $(b)$... i have seen you have posted this as a question I do not understand your idea of posting a question and then writing the same thing as an asnwer... which one do you prefer to remove? This answer or that question? That question is giving you nothing more than a negative vote.. $\endgroup$
    – user87543
    Jan 23, 2014 at 11:27
  • $\begingroup$ @PraphullaKoushik .. ha ha ha .. negative votes doesn't matter as long as I am learning mathematics. Thanks for supporting my question though.. I didn't see your question after someone flagged my question as duplicate.. Since I didn't get anything there I posted here hoping that you would suggest something.. And you did.. $\endgroup$ Jan 23, 2014 at 15:45
  • $\begingroup$ Let us be practical... I guess negative votes does matter.. you will get frustrated in no time if you keep getting negative votes for every thing you do... so, please come out of "negative votes doesn't matter as long as I am learning mathematics. " It would be good for you... Good luck! $\endgroup$
    – user87543
    Jan 24, 2014 at 1:30

Comment on your first case:

$n^{1.5}a_n$ can converge to 1.5 while always staying above 1.5. But it must eventually go below (say) 3 so that eventually $a_n \leq 3/n^{1.5}$.

Comment on your third case:

You wrote since $a_n \geq 1/n^2$ (which isn't true anyway), comparison test applies which is also false. What you really get is $a_n \leq a_1/n^2$ (easily verified) which is enough to give convergence.

  • $\begingroup$ For third case that was actually a typo (:D) I was meant to say $a_n\leq \frac{1}{n^2}$ do you think saying $a_n\leq \frac{1}{n^2}$ is not sufficient? for first case I do understand that though $\frac{1}{n} > 0$ we would get $\frac{1}{n}$ converging to $0$ similarly there might be a chance that "$n^{\frac{3}{2}}a_n$ can converge to $\frac{3}{2}$ while always staying above $\frac{3}{2}$" but then i could not understand your next statement "But it must eventually go below (say) $3$ so that eventually $a_n≤3/n^{1.5}$." could you please explain a bit more about this,,, $\endgroup$
    – user87543
    Jan 17, 2014 at 2:17
  • 1
    $\begingroup$ I can only tell you to re-read your solution again and again until you find mistakes. If you cannot, just move on. $\endgroup$
    – hot_queen
    Jan 17, 2014 at 4:49

You must log in to answer this question.