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for example: I have a series enter image description here

is there numerical computation method to find it ? thanks

T.Nhan
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    If you say numerical, then just add the terms until you meet the convergence criteria (there are acceleration techniques). May be, you ask for the exact closed form expression of such monsters ? – Claude Leibovici Mar 28 '15 at 15:24
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    Hint: Read this article, presenting Euler's intuitive approach to the Basel problem, then differentiate the natural logarithm of the infinite product expression for $\dfrac{\sin\pi x}{\pi x}~,~$ and, after decomposing the general term of your infinite series into partial fractions, use it to evaluate it. – Lucian Mar 28 '15 at 15:28
  • @Lucian. With this, you make my day ! Thanks. – Claude Leibovici Mar 28 '15 at 15:33
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    If you are lazy, you can throw this to WA which will use essentially what mentioned in Lucian's comment and evaluate your sum symbolically to $\frac{1}{800}\left(54\pi\coth\left(\frac{\pi}{6}\right)+5\pi^2-400\right)$ – achille hui Mar 28 '15 at 15:46

2 Answers2

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Your terms are proportional to $\frac1{16·36·n^6}$, accordingly the error of the n-th partial sum is about $\frac1{16·36·5·n^5}$, so the first 100 terms, summed from smallest to largest, should give a valid numerical result.

You might also try out the Epsilon algorithm of P. Wynn, "On the Convergence and Stability of the Epsilon Algorithm", SIAM Journal on Numerical Analysis 3 (1) (Mar 1966), p. 91–122

Lutz Lehmann
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Yoy can write $$(1 - 4 n^2)^2 (36 n^2 + 1)=(2 n-1)^2 (2 n+1)^2 (6 n-i) (6 n+i)$$ Using partiai fractions, the summand is $$\frac{7}{100 (2 n+1)}-\frac{7}{100 (2 n-1)}+\frac{81 i}{200 (6 n+i)}-\frac{81 i}{200 (6 n-i)}+$$ $$\frac{1}{40 (2 n+1)^2}+\frac{1}{40 (2 n-1)^2}$$

Computing the partial sums up to $p$, we have at the end for $800 S_p$ $$\frac{56}{2 p+1}+\frac{5}{\left(p+\frac{1}{2}\right)^2}-54 i \psi \left(p+\left(1-\frac{i}{6}\right)\right)+54 i \psi \left(p+\left(1+\frac{i}{6}\right)\right)-10 \psi ^{(1)}\left(p+\frac{1}{2}\right)+5 \pi ^2-400+54 \pi \coth \left(\frac{\pi }{6}\right)$$

Now, using asymptotics and Taylor expansions $$S_p=\frac{1}{800} \left(5 \pi ^2+54 \pi \coth \left(\frac{\pi }{6}\right)-400\right)-\frac{1}{2880 p^5}+\frac{1}{1152 p^6}+O\left(\frac{1}{p^7}\right)$$

Computing for $p=10$ $$S_{10}=\frac{89936259827237312633515635583570094}{29615662049611209795735649868626520625}$$ which is $0.00303678032$ while the above truncated expansion gives

$$S_{10}\sim -\frac{192000001}{384000000}+\frac{\pi ^2}{160}+\frac{27}{400} \pi \coth \left(\frac{\pi }{6}\right)$$ which is $0.00303678042$

Claude Leibovici
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