6
$\begingroup$

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).

Thanks!

  • 0
    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).2012-02-04

3 Answers 3

12

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 $x \to \infty$.

8

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.

2

$ \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.$