My problem is: What is the expression in $n$ that equals to $\sum_{i=1}^n \frac{1}{i^2}$?
Thank you very much~
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2There isn't one. – TonyK Dec 16 '10 at 14:31
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Thank you very much. Do you mean algebraic expression in $n$? I guess there is no algebraic expression in $n$ that satisfies the equation. But I am wondering if there is a transcendetal expression in $n$ that does. – ShinyaSakai Dec 18 '10 at 18:35
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I don't think there is a "closed" form. You can give a good approximation using the Euler-McLaurin Summation formula though:
$$\sum_{j=1}^{n} \dfrac{1}{j^2} = \dfrac{\pi^2}{6} - \dfrac{1}{n} - \dfrac{1}{2n^2} + \mathcal{O}(\dfrac{1}{n^3})$$
(If you need more accuracy you can include more terms from the summation formula to give the coefficients of the lower order terms)
Note: The Euler McLaurin Summation formula only tells us that
$$\sum_{j=1}^{n} \dfrac{1}{j^2} = C - \dfrac{1}{n} - \dfrac{1}{2n^2} + \mathcal{O}(\dfrac{1}{n^3})$$
for some constant $\displaystyle C$.
We know by other means that $\displaystyle C = \dfrac{\pi^2}{6}$, for instance, see this for a multitude of ways: Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$
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I not sure of the thrust of your question but maybe the generalised harmonic numbers are what you want $$ H_{n,r} = \sum_{k=1}^n \frac{1}{k^r} , $$
and in particular $H_{n,2}$
You can find more information here, including a very nice identity for $H_{n,2}$ by B. Cloitre.
Derek Jennings
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1Alternatively, you have the expression in terms of the trigamma function: $\frac{\pi^2}{6}-\psi^{(1)}(n+1)$ – J. M. ain't a mathematician Dec 16 '10 at 14:57
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1I am guessing OP is looking for a simpler formula than just $\sum \dfrac{1}{j^2}$. I suppose something similar to $\sum j = n(n+1)/2$. – Aryabhata Dec 16 '10 at 15:18
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2@Moron: Very likely, let's hope he won't be too disappointed :-) – Derek Jennings Dec 16 '10 at 15:36
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@Mo: Well, Derek and me gave it as simple as it can be... again, what exactly is a "closed form"? ;) – J. M. ain't a mathematician Dec 16 '10 at 15:45
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1@Derek: what you are proposing is that we find a closed form for the harmonic numbers by defining 'closed form' to include harmonic numbers! :) – Mariano Suárez-Álvarez Dec 16 '10 at 16:24
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@J.M: Your's is simple, but kind of circular, perhaps? I won't call the identity given by Derek to be simpler :-) I agree, though. – Aryabhata Dec 16 '10 at 16:37
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As I said in a previous post @Mo, polygamma functions and (generalized) harmonic numbers are virtually the same. :) – J. M. ain't a mathematician Dec 17 '10 at 00:07
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@J.M: Yep, I agree. In fact I realized that the circularity isn't there after I posted (perhaps you can add an answer?). btw, you need at least 3 letter to be able to notify :-) – Aryabhata Dec 17 '10 at 01:40
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But I don't like calling you @Moron. :P Anyway Mo, I'd rather Derek edit his answer instead; I'll only be effectively copying Derek's answer otherwise, except that the notation is different. – J. M. ain't a mathematician Dec 17 '10 at 02:03
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@J.M: :-P. Either way, as long as we have some answer which mentions that. I think it is too relevant to be left as a comment. – Aryabhata Dec 17 '10 at 02:14
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Thanks, everyone~ In fact, as Moron San have said, I was looking for some "simple" expression. But I really learn much from this answer~ – ShinyaSakai Dec 18 '10 at 18:26
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I am not sure if this will work or not, but maybe you could try writing the expression in terms of falling factorials. Then maybe use summation by parts. I am not sure how nicely this will work, but you could try it. Let me know what you find out!
FUZxxl
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Tyler Clark
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