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I can prove this sum has a constant upper bound like this:

$$\sum_{k=1}^n \frac1{k ^ 2} \lt 1 + \sum_{k=2}^n \frac 1 {k (k - 1)} = 2 - \frac 1 n \lt 2$$

And computer calculation shows that sum seems to converge to 1.6449. But I still want to know:

  • Dose this sum converge?
  • Is there a name of this sum (or the series $1 / k ^2 $)?
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6  
Converge? Well, sure... –  J. M. Aug 22 '11 at 3:00
    
This is a special case of a $p$-series. $\sum{1/k^p}$ converges if and only if $p\gt 1$. –  Arturo Magidin Aug 22 '11 at 3:04
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This cute general proof by joriki should interest you as well. –  J. M. Aug 22 '11 at 3:10
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Arturo's comment reminded me of something that had been bugging me lately: it's already known as a $\zeta$ function, so why did the name "p-series" have to be given to it as well (unless that was how it was called before Riemann's investigations...)? –  J. M. Aug 22 '11 at 3:16
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A nice and very elementary proof is contained in the following preprint of Dr Daniel Daners (Univeristy of Sydney). It uses tools that most high school students can understand. maths.usyd.edu.au/u/daners/publ/abstracts/zeta2/zeta2.pdf –  Ragib Zaman Aug 22 '11 at 3:23

2 Answers 2

up vote 9 down vote accepted

A sequence that is increasing and bounded must converge. That's one of the fundamental properties of the real line. So once you've observed that your sequence of partial sums is bounded, since it obviously increases, it must converge. Of course it is a very famous series, and it converges to a number which quite miraculously has a "closed form" formula: it is $\pi^2/6$.

EDIT: for many proofs of this famous formula, see this MO question.

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Yes, it does converge, and believe it or not, $$\sum_{k=1}^\infty\frac{1}{k^2}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k^2}=\frac{\pi^2}{6}$$ Determining the specific value of this infinite sum was originally known as the Basel problem, and Euler was the first person to determine the correct value of the sum, although his initial methods were not 100% rigorous (but they can be made to be rigorous, and zyx points out below that Euler later gave valid proofs).

In general, the infinite sum $$\sum_{k=1}^\infty\frac{1}{k^s}$$ is extremely important in number theory, so much so that we give it a name, the Riemann zeta function, $\zeta(s)$. That is, $$\zeta(s)=\sum_{k=1}^\infty\frac{1}{k^s}.$$ So, we see that $\zeta(2)=\frac{\pi^2}{6}$. Wikipedia lists some other known values of the zeta function. A priori, the function $\zeta(s)$ is defined for any real number $s>1$ (that is, the sum $\sum_{k=1}^\infty\frac{1}{k^s}$ will converge for any $s>1$); but in fact, there is a well-defined notion of raising integers to complex numbers, and then we get that $\zeta(s)$ is well-defined for every $s\in\mathbb{C}$ for which $\text{Re}(s)>1$. Using something called analytic continuation, we can then define $\zeta(s)$ for every complex number (other than $s=1$).

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Euler gave several different computations of zeta(2) over a period of decades. His first computation of zeta(2) was not completely rigorous but some of the later proofs are correct. He also developed methods of numerical calculation to (rigorously) estimate the sum more accurately than was possible by simple hand calculation of the first $N$ terms. –  zyx Aug 22 '11 at 3:25
    
@zyx: Thank you for the explanation, I've incorporated it into my post. I wouldn't want to denigrate Euler :) –  Zev Chonoles Aug 22 '11 at 3:27
    
"He also developed methods of numerical calculation..." - that'd be what we now call the "Euler-Maclaurin formula". –  J. M. Aug 22 '11 at 3:30
    
@J.M.: and acceleration of series, and ... –  zyx Aug 22 '11 at 5:03
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@J.M., prior to the Euler MacLaurin formula, Euler computed zeta(2) to six decimal places using a particular acceleration of the series. E-M formula added much more accuracy, and the ability to approximate zeta(n) in general. The derivation of the exact zeta(2) came several years later. The numerical methods Euler applied to (or developed for the very purpose of calculating) zeta(2) went beyond the E-M formula, both before and after the discovery of the exact evaluation. –  zyx Aug 22 '11 at 8:34

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