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I want to ask you if can it be so simple to prove that $\lim _{x \to \infty}\sum_{1}^{\infty}\frac{x^2}{1+n^2x^2}=\sum_{1}^{\infty}\frac{1}{n^2}$ by divide the numerator and denominator with $x^2$ and that's it?

If it this simple indeed you can write a comment and I'll delete the question after I'll read it, or perhaps I'm missing something important (and I should involve power series).


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I think it is not enough. For any $x$, the sum with the $x$ stuff is certainly less than $\sum 1/n^2$. But we need to show that for $x$ large, the sum is very little less. So I would take the difference between $1/n^2$ and the same term with the $x$ stuff, and show that for $x$ large the sum of these differences is small. Details should not take too long! I don't see as simple an argument using power series. – André Nicolas Feb 3 '12 at 18:52
I suggest you check… – Pedro Tamaroff Feb 4 '12 at 0:47
It is always very tempting to want to interchange the series limit and the limit on $x$, but we have to resist the temptation! It is very important to consider uniform convergence (which is one of the main properties that makes power series so very awesome). – Aru Ray Feb 4 '12 at 2:32
up vote 12 down vote accepted

Use $$ \frac{1}{n^2} \frac{x^2}{1+x^2} \leqslant \frac{1}{n^2} \frac{x^2}{x^2 + n^{-2}} < \frac{1}{n^2} $$ Thus $$ \frac{x^2}{1+x^2} \sum_{n=1}^\infty \frac{1}{n^2} \leqslant \sum_{n=1}^\infty \frac{x^2}{n^2+x^2} < \sum_{n=1}^\infty \frac{1}{n^2} $$ Both upper and the lower bounds have the same limit as $x \to \infty$.

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In general, interchanging limits and sums or integrals can be tricky. It's not always true that
$\lim_{x \to \infty} \sum_{n=1}^\infty f_n(x) = \sum_{n=1}^\infty \lim_{x \to \infty} f_n(x)$, even when both sides converge. It is true for dominated convergence (if there is some convergent series $\sum_{n=1}^\infty b_n$ with $|f_n(x)| \le b_n$ for all $n$ and $x$) and monotone convergence (if $f_n(x)$ is positive, and increasing as a function of $x$). This example satisfies both conditions.

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$$ \sum \left( \frac{1}{n^2} - \frac{x^2}{1+n^2x^2} \right) = \sum \frac{ 1}{n^2(1+n^2 x^2) } \leq \frac{1}{x^2} \sum \frac{1}{n^4} \to 0.$$

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