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I'm investigating the Basel Problem, and the sum to consider is:

$\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{n^2}$

How can I show this converges? Using graphs/computer software is also fine, but how would I do it? Is there a way using calculus?

I've looked on Google for a while and I found a few proofs, but I didn't understand any of them.

Jim
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3 Answers3

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Maybe the simplest way of proving this is to use :

$ \frac{1}{n^2} \leq \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n} $ for n $\geq 2$

Introducing partial sums you get :

$ S_n = \sum_{k=2}^n \frac{1}{k^2} \leq \sum_{k=2}^n \frac{1}{k-1} - \frac{1}{k} =1 - \frac{1}{n} \leq 1 $

So $(S_n)$ is an increasing sequence, upper-bounded hence...

Edit: I'll write down the part about fractions since you're not very familiar with it :) .

You can deal with this in two ways:

1) Write that: $\frac{1}{n(n-1)} = \frac{A}{n} +\frac{B}{n-1} $, then reduce to a common denominator ,i.e :

$\frac{A}{n} +\frac{B}{n-1} = \frac{A(n-1) + Bn}{n(n-1)}$ , then identify with the numerator: $-1=A ; A+B=0 \implies A=-1; B=1$

2) Directly verifying: $\frac{1}{n-1} - \frac{1}{n} = \frac{n}{n(n-1)} - \frac{n-1}{(n-1)n} =\frac{1}{n(n-1)} $

Here the first one sounds silly because the result can be easily obtained by finding the A and B directly, that is reducing $\frac{1}{n-1} - \frac{1}{n}$ to a common denominator and verify the equality. However this method of expanding a fraction or reducing it, depending on what we are supposed to prove, is very usual. For further information about that, I suggest you take a look a this page:

http://en.wikipedia.org/wiki/Partial_fraction_decomposition You'll find everything you need to know. It explains why I can write the decomposition with A and B and expect to find something that has a meaning. It's not always the case.

As for bounding the sum, the latter introduces a telescoping with the sequence:

$V_n = \frac{1}{n}$

$ \frac{1}{k^2} \leq \frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k} = V_{k-1} -V_k $

$ \implies \sum_{k=2}^n \frac{1}{k^2} \leq \sum_{k=2}^n (V_{k-1} -V_k) = (V_1-V_2)+(V_2-V_3)+...+(V_{n-2}-V_{n-1})+(V_{n-1}-V_n) = V_1 - V_n = 1-\frac{1}{n} $

Only the first and last terms remain.

mvggz
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    This is also how Bernoulli proved the convergence. – Math-fun Nov 26 '14 at 14:48
  • I think this is one of the proof that uses the most basic notions, thereby being one the easiest to understand :) – mvggz Nov 26 '14 at 14:52
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    $S_n : = : 1+... ;;;$ –  Nov 26 '14 at 16:32
  • @Ricky Demer I'm not sure of what you meant by that, since the partial starts at k=2..If you want to start it a k=1, the upper bound becomes : $2-\frac{1}{n}$ – mvggz Nov 27 '14 at 14:30
  • Hi, I've used Julian Aguirre's method but I'd also like to explore this method in my work. Please can you explain this in a bit more detail? I don't understand most of it. – Jim Dec 04 '14 at 11:24
  • @Jim yes of course. But I will need to know your background in terms of knowledge/practice so that I can understand what's troubling you in the proof. Here I use very basic notions to make it the most understandable, merely one property about real valued sequences:
    • If ($S_n$) is a positive sequence (ie for all n, $S_n \geq 0$), bounded and increasing (ie for all n, $S_{n} \leq S_{n+1} \leq S$), then ($S_n$) converges. This won't give you the value of the limit, only the certainty of its existence.

    Now what is bothering you?

     – mvggz Dec 04 '14 at 13:30
  • @Jim, or is it the way I've bounded ($S_n$)? If so, note that : $n \geq n-1 \implies \frac{1}{n} \leq \frac{1}{n-1} \implies \frac{1}{nn} \leq \frac{1}{(n-1)n}$ – mvggz Dec 04 '14 at 13:32
  • @mvggz could you explain the partial fraction bit when bounding? I haven't actually learnt about them formally and I don't really understand the explanations online. I have:

    $\dfrac{1}{n(n-1)}=\dfrac{A}{n-1}+\dfrac{B}{n}$ Is this correct? If so how do I show that $A=1, B=-1$?

     – Jim Dec 08 '14 at 11:46
  • @Jim, yes it is correct. I'll edit my answer to detail this to you – mvggz Dec 08 '14 at 12:13
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Note that $$ \begin{alignat}{8} &\frac{1}{1^2}+&&\frac{1}{2^2}+&&\frac{1}{3^2}+&&\frac{1}{4^2}+&&\frac{1}{5^2}+&&\frac{1}{6^2}+&&\frac{1}{7^2}+&&\dots\\ \leq &\frac{1}{1^2}+&&\frac{1}{2^2}+&&\frac{1}{2^2}+&&\frac{1}{4^2}+&&\frac{1}{4^2}+&&\frac{1}{4^2}+&&\frac{1}{4^2}+&&\dots \end{alignat} $$ where there are $2^n$ copies of $\frac{1}{(2^n)^2}$.

But $\frac{1}{(2^n)^2}=\frac{1}{4^n}$, so $2^n$ copies of it is $\frac{2^n}{4^n}=\frac{1}{2^n}$. So the sum is bounded above by $\sum_{k=0}^\infty \frac{1}{2^n}=2$.

Micah
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Another way of proving it that generalizes to sums of the type $\sum f(n)$ with $f$ decreasing: $$ \sum_{n=1}^N\frac{1}{n^2}\le1+\int_1^N\frac{dx}{x^2}. $$

Julián Aguirre
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