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*}
Evaluating a limit, $\lim\limits_{t\rightarrow 0^{+}} {\sum\limits_{n=1} ^{\infty} \frac{\sqrt{t}}{1+tn^2}}$
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calculus
limits
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3@Patrick: I beg to differ. – 2011-08-17
2 Answers
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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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0@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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3@J.M., this solution is eye-opening for me. Thanks! – 2011-08-17