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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*}

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    $\begingroup$ 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. $\endgroup$ – J. M. is a poor mathematician Aug 17 '11 at 5:13
  • $\begingroup$ 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. $\endgroup$ – Zev Chonoles Aug 17 '11 at 5:19
  • $\begingroup$ @J.M. : This is not very satisfying... $\endgroup$ – Patrick Da Silva Aug 17 '11 at 5:20
  • $\begingroup$ 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. $\endgroup$ – J. M. is a poor mathematician Aug 17 '11 at 5:25
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    $\begingroup$ @Patrick: I beg to differ. $\endgroup$ – anon Aug 17 '11 at 5:47
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Hint: $$ \sqrt t \int_1^\infty {\frac{1}{{1 + tx^2 }}\,dx} = \int_{\sqrt t }^\infty {\frac{1}{{1 + x^2 }}\,dx} \to \frac{\pi }{2}. $$

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  • $\begingroup$ @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 $\sqrt{t}$. $\endgroup$ – Syang Chen Aug 17 '11 at 14:06
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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 $\pi/2$.

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    $\begingroup$ J.M.: Seems my only real contribution was noticing $\coth \infty = 1$ ;) $\endgroup$ – anon Aug 17 '11 at 5:57
  • $\begingroup$ In my defense, I was hungry while writing the comment to the question, and rumbling stomachs are distracting... :) $\endgroup$ – J. M. is a poor mathematician Aug 17 '11 at 5:59
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    $\begingroup$ @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... $\endgroup$ – J. M. is a poor mathematician Aug 17 '11 at 6:47
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    $\begingroup$ @J.M., this solution is eye-opening for me. Thanks! $\endgroup$ – Syang Chen Aug 17 '11 at 13:56

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