What is the expression of $n$ that equals to $\sum_{i=1}^n \frac{1}{i^2}$?

Question

My problem is: What is the expression in $n$ that equals to $\sum_{i=1}^n \frac{1}{i^2}$?

Thank you very much~

Answer 1

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 $\sum_{n=1}^\infty \frac{1}{n^2}$.

Answer 2

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.

Answer 3

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!