Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It seems that the following limit exists. But I couldn't figure out the exact value. Anyone could help me? Thanks! \begin{align*} \lim_{t\rightarrow 0^{+}} {\sum_{n=1} ^{\infty} \frac{\sqrt{t}}{1+tn^2}} \end{align*}

share|cite|improve this question
I was about to write an answer... anyway, your series can be derived from the partial fraction series for the hyperbolic cotangent, which leads to the limit $$\frac12\lim_{t\to 0^+} \left(\pi\coth\frac{\pi}{\sqrt t}-\sqrt t\right)$$ ... on the other hand, $\pi/2$ is the answer that I seem to be getting. – J. M. Aug 17 '11 at 5:13
Ah, you're right J.M.; I don't think we can interchange the sum and limit here, since $\frac{\sqrt{t}}{1+tn^2}$ decays too slowly. – Zev Chonoles Aug 17 '11 at 5:19
@J.M. : This is not very satisfying... – Patrick Da Silva Aug 17 '11 at 5:20
Right, so I haven't written an answer yet @Patrick, and merely left it as a comment. I have to go to my lunch now, and maybe finish this later unless somebody can fill the gaps in my idea. – J. M. Aug 17 '11 at 5:25
@Patrick: I beg to differ. – anon Aug 17 '11 at 5:47
up vote 11 down vote accepted

Hint: $$ \sqrt t \int_1^\infty {\frac{1}{{1 + tx^2 }}\,dx} = \int_{\sqrt t }^\infty {\frac{1}{{1 + x^2 }}\,dx} \to \frac{\pi }{2}. $$

share|cite|improve this answer
Nice! $\text{}$ – J. M. Aug 17 '11 at 6:06
@J. M. - Thanks. – Shai Covo Aug 17 '11 at 6:08
Like a bawss. +1, and OP should definitely accept this one. – Patrick Da Silva Aug 17 '11 at 6:30
@Patrick - Thanks. – Shai Covo Aug 17 '11 at 6:33
@Shai, thank you. I tried this before I asked the question. But I failed to see that $\lim\limits_{t\rightarrow 0^{+}} |{\sum\limits_{n=1} ^{\infty} \frac{\sqrt{t}}{1+tn^2}}-\int\limits_0^\infty {\frac{\sqrt t }{{1 + tx^2 }}\,dx}|=0$. Now I could see that the difference is actually controlled by $\sqrt{t}$. – Syang Chen Aug 17 '11 at 14:06

Note the hyperbolic cotangent identity

$$\sum_{n=1}^\infty\frac{1}{z^2+n^2} =\frac{\pi z \coth(\pi z)-1}{2z^2}.$$

Replace $t$ with $t^2$ for convenience. Then

$$\sum_{n=1}^\infty\frac{t}{1+t^2n^2} =\frac{1}{t}\sum_{n=1}^\infty \frac{1}{t^{-2}+n^2}=\frac{1}{2} \left[\pi\coth(\pi/t)-t\right].$$

Observe that $\coth(s)\to1$ as $s\to\infty$ and $t\to0$, so the limit is $\pi/2$.

share|cite|improve this answer
Thanks for finishing! – J. M. Aug 17 '11 at 5:51
J.M.: Seems my only real contribution was noticing $\coth \infty = 1$ ;) – anon Aug 17 '11 at 5:57
@Patrick: proving that partial fraction decomposition in here would require a bit of length, unfortunately... :P You might enjoy this; it should be nicely readable... – J. M. Aug 17 '11 at 6:47
@J.M., this solution is eye-opening for me. Thanks! – Syang Chen Aug 17 '11 at 13:56
"random shitty identity" - ??? Well, 2 out of 3 ain't bad...$$$$ :) – The Chaz 2.0 Aug 17 '11 at 18:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.