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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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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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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