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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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    $\begingroup$ There isn't one. $\endgroup$ – TonyK Dec 16 '10 at 14:31
  • $\begingroup$ 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. $\endgroup$ – 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.

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    $\begingroup$ Alternatively, you have the expression in terms of the trigamma function: $\frac{\pi^2}{6}-\psi^{(1)}(n+1)$ $\endgroup$ – J. M. isn't a mathematician Dec 16 '10 at 14:57
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    $\begingroup$ I am guessing OP is looking for a simpler formula than just $\sum \dfrac{1}{j^2}$. I suppose something similar to $\sum j = n(n+1)/2$. $\endgroup$ – Aryabhata Dec 16 '10 at 15:18
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    $\begingroup$ @Moron: Very likely, let's hope he won't be too disappointed :-) $\endgroup$ – Derek Jennings Dec 16 '10 at 15:36
  • $\begingroup$ @Mo: Well, Derek and me gave it as simple as it can be... again, what exactly is a "closed form"? ;) $\endgroup$ – J. M. isn't a mathematician Dec 16 '10 at 15:45
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    $\begingroup$ @Derek: what you are proposing is that we find a closed form for the harmonic numbers by defining 'closed form' to include harmonic numbers! :) $\endgroup$ – Mariano Suárez-Álvarez Dec 16 '10 at 16:24
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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!

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    $\begingroup$ Thanks~ I will take time to have a closer look to this~ $\endgroup$ – ShinyaSakai Dec 18 '10 at 18:31
  • $\begingroup$ @ShinyaSakai Not a problem. Let me know what you find. $\endgroup$ – Tyler Clark Dec 19 '10 at 22:34

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