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What is the value of the following finite series $$\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\cdots+\dfrac{1}{n^2}~=~?$$

  • I know that the value of the infinite series $$\sum_{i=1}^\infty\dfrac 1{i^2}=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\cdots+\dfrac{1}{n^2}+\cdots~=~\dfrac{~~\pi^2}{6}~.$$But I am unable to find the value of the series when the number of terms are finite $($say,$~n~)$.

Please help.

nmasanta
  • 9,640

2 Answers2

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This is given by

$$\sum_{k=1}^n\frac1{k^2}=\zeta(2)-\zeta(2,n+1)=\zeta(2)-\psi^{(1)}(n+1)$$

in terms of the Hurwitz zeta function and trigamma function. Approximations and other forms can be found in the provided links.

Simply Beautiful Art
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You can write $$S_n=\sum_{k=2}^n\frac1{k^2}=H_n^{(2)}-1$$ and, for large values of $n$ $$S_n=\left(\frac{\pi ^2}{6}-1\right)-\frac{1}{n}+\frac{1}{2 n^2}-\frac{1}{6 n^3}+\frac{1}{2 n^4}-\frac{29}{30 n^5}+\frac{3}{2 n^6}+O\left(\frac{1}{n^7}\right)$$ For $n=10$, the exact value is $$S_{10}=\frac{698249}{1270080}\approx 0.549767$$ while the above gives $$\frac{\pi ^2}{6}-\frac{6570749}{6000000}\approx 0.549809$$

Claude Leibovici
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