I'm having trouble finding the limit of an infinite series.

The series is: $$f(x)=\sum_{n=0}^\infty \frac{1}{1+n^2x}$$ on $(0, \infty)$. I have to prove:

1) $\lim \limits_{x \to \infty}f(x) = 1$

2) $\lim \limits_{x \to 0^{+}}f(x) = \infty$

How do I go about manipulating the sequence so that I can take a limit? Any help/hints would be appreciated.

  • 1
    $\begingroup$ $\displaystyle\sum_{n=-\infty }^\infty\frac1{1+n^2t^2} ~=~ \frac\pi t~\coth\frac\pi t~.~$ This can be proven by differentiating the natural logarithm of Euler's infinite product formula for the sine function. $\endgroup$
    – Lucian
    Nov 4 '15 at 2:26

1) Forget temporarily about the first term. For $n\ge 1$, and positive $x$, the $n$-th term is positive and $\lt \frac{1}{x}\cdot \frac{1}{n^2}$.

So the sum of all the terms except the first is therefore $\lt \frac{1}{x}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdot\right)$. The series $\sum_1^\infty \frac{1}{k^2}$ converges.

2) Suppose that $0\lt x\le \frac{1}{N^2}$. Then each term up to the term $\frac{1}{1+xN^2}$ is $\ge \frac{1}{2}$. So the partial sum up to this term is $\ge \frac{N+1}{2}$.


For a moment, forget the first term. By the integral formula for approximation, since the function is positive and decreasing,


Since we know that: $$\int\frac{1}{1+xn^2}dn=\frac{1}{\sqrt{x}}\tan^{-1}(\sqrt{x}n)$$ And: $$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{x}}\tan^{-1}(\sqrt{x}n)=\frac{\pi}{2\sqrt{x}}$$ Therefore $$\frac{1}{2\sqrt{x}}-\tan^{-1}\sqrt{x}\leq\sum_{n=1}^\infty\frac{1}{1+xn^2}\leq\frac{1}{2\sqrt{x}}$$ So finally, $$\frac{1}{2\sqrt{x}}-\tan^{-1}\sqrt{x}+1\leq\sum_{n=0}^\infty\frac{1}{1+xn^2}\leq\frac{1}{2\sqrt{x}}+1$$ Analyzing both limits at infinity, then: $$1-\frac{\pi}{2}\leq\sum_{n=0}^\infty\frac{1}{1+xn^2}\leq1$$

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