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Could you help me, the task is to determine the number of terms needed to sum up to reach accuracy of 5 decimal places. The series is: $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$

I have seen that $\sum_{n=1}^{\infty}{\frac{1}{n^2}}=\frac{\pi^2}{6}$, but this isn't excatly what I want but may be this will useful somehow, I don't have any good ideas how to start

J. W. Tanner
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Artkol
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2 Answers2

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The error is the sum of the items that you didn’t add up. For example summing up to m, you didn’t add up 1/n^2 from m+1 upwards.

The error is the sum of 1/n^2 from m+1 to infinity, which is about the integral of 1/n^2, which is about -1/n from m to infinity, which is about 1/m.

So for five decimals you need to add up 100,000 terms. For say 10 digits you would have to be careful if you add the sum using floating-point arithmetic because of significant rounding errors.

gnasher729
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    Maybe you could loose all the "abouts" in your second paragraph... we just need an upper bound on the error. $$\sum_{n=m+1}^{+\infty} \frac{1}{n^2}< \int_m^{+\infty} \frac{1}{x^2} dx = \frac 1m$$ – PierreCarre Dec 18 '24 at 12:33
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I think you need to use programming to find $n$ such that condition satisfy. But as a help I made a simple desmos for you ,to find idea

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enter image description here

Khosrotash
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    You do not need any programing to show that $$\sum_{k+1}^\infty\frac 1{n^2}=\psi ^{(1)}(k+1)=\frac{1}{k}-\frac{1}{2 k^2}+O\left(\frac{1}{k^3}\right)$$ – Claude Leibovici Dec 18 '24 at 08:17
  • It is better if you get an error bound that does not depend on knowing the exact solution. Having the exact solution renders the exercise of computing partial sums useless. – PierreCarre Dec 18 '24 at 12:25