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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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    @Patrick: I beg to differ.2011-08-17

2 Answers 2

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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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    @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 $\sqrt{t}$.2011-08-17
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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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    @J.M., this solution is eye-opening for me. Thanks!2011-08-17