Proving $\frac{1}{n^2}$ infinite series converges without integral test [duplicate]

Just out of curiosity, I was wondering if anybody knows any methods (other than the integral test) of proving the infinite series where the nth term is given by $\frac{1}{n^2}$ converges.

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marked as duplicate by Ayman Hourieh, T. Bongers, Shuhao Cao, Amzoti, Julian KuelshammerAug 22 '13 at 20:45

–  Argon Aug 22 '13 at 18:42
Cauchy Condensation. Works for $\sum \frac{1}{n^p}$, $p\gt 1$. –  André Nicolas Aug 22 '13 at 19:01

Hint: $$\frac{1}{n^2} < \frac{1}{n(n-1)}.$$

$$\frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 2} + \ldots + \frac{1}{n(n-1)} = 1 - \frac12 + \frac12 - \frac13 + \ldots + \frac{1}{n-1}-\frac1n = 1 - \frac1n \to 1.$$

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I like this solution very much. –  Jack Mapleton Aug 22 '13 at 18:44
I'm missing something... how do you end up with that expansion? –  dfeuer Aug 22 '13 at 18:45
He did a partial fraction decomposition. –  Jack Mapleton Aug 22 '13 at 18:46
@dfeuer, $\frac{1}{n(n-1)} = \frac{n-(n-1)}{n(n-1)}$. –  njguliyev Aug 22 '13 at 18:47
Ah, thanks. Nice telescope, Galileo. –  dfeuer Aug 22 '13 at 18:56

With less words. Hopefully clear enough. Oresme's style, but converging this time, and proving that the sum is $<2$.

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Do you happen to have a source for this? I'd like to see others like this if there is some central repository... –  daniel Aug 22 '13 at 19:50
I've seen it in some calculus book. I reproduced this with Mathematica. –  Jyrki Lahtonen Aug 22 '13 at 19:50
I guess it would look neater with narrower borders. –  Pedro Tamaroff Aug 22 '13 at 20:00
May be, @Peter. First I had no borders at all, then very thin ones, then way too thick. This is a compromise. I agree with you in the sense that the next stack would look horrible with this border thickness. Too bad I already closed the notebook. Also, there would be more orphaned images at imgur :-/ –  Jyrki Lahtonen Aug 22 '13 at 20:03
This is a very good illustration of the cauchy's condensation test that for a positive non-increasing function $f(n)$ the series $\sum f(n)$ and $\sum 2^{n}f(2^{n})$ converge/diverge together. –  Paramanand Singh Oct 5 '13 at 10:44