Limit of infinite summation 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.
 A: 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 $\ge \frac{N+1}{2}$. 
A: For a moment, forget the first term. By the integral formula for approximation, since the function is positive and decreasing,
$$\int_1^\infty\frac{1}{1+xn^2}dn\leq\sum_{n=1}^\infty\frac{1}{1+xn^2}\leq\int_0^\infty\frac{1}{1+xn^2}dn$$
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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