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How to find the summation of the above series up to infinity? I have tried to find the above summation by putting n=0,1,2,3,...and so on. Finally I got the above series to be (1-(1/2)+(1/23)-(1/24)+(1/45)-(1/46)+...) and so on. But I don't know how to proceed further because this is not even a telescopic series. Please help me out to find the summation up to infinity.

J. W. Tanner
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DEEPANWITA Roy
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    Well possible that there is only a numerical value. Giving it to Wolfram alpha is a good trial. – Peter Sep 18 '23 at 16:19
  • But is there any formula to find the summation upto infinity. – DEEPANWITA Roy Sep 18 '23 at 16:21
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    Note that the bracketed term is equal to $1/[(1+22n)(2+22n)]$, though I'm not certain this leads anywhere aside from an easy proof that the series converges. WolframAlpha gives the sum in terms of the digamma function. If this is from a problem set, did the problem ask to find the value of the sum or just show it converges? – eyeballfrog Sep 18 '23 at 16:26
  • @eyeballfrog the question was asked to find the summation. – DEEPANWITA Roy Sep 18 '23 at 16:32
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    A strange exercise since there is apparently no nice formula. – Peter Sep 18 '23 at 16:39
  • Yes@Peter. You are correct. – DEEPANWITA Roy Sep 18 '23 at 16:45
  • It seems WolframAlpha sometimes gives the answer as a pile of logs and trig functions rather than the digamma function. These expressions are in fact equal, as the digamma function has closed form for all rational numbers. Not necessarily a nice closed form, but closed nonetheless. – eyeballfrog Sep 18 '23 at 16:55
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    But how should this be found without electronic help ? Entering it in Wolfram Alpha cannot be the desired solution. – Peter Sep 18 '23 at 17:07
  • You are right@Peter. – DEEPANWITA Roy Sep 18 '23 at 18:04

1 Answers1

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Consider a partial sum $$S_p(a,b)=\sum_{n=0}^p\frac 1{a n+b}=\frac 1 a\left(\psi \left(p+1+\frac{b}{a}\right)-\psi \left(\frac{b}{a}\right)\right)$$ For large values of $p$ $$S_p(a,b)=\frac{\log (p)-\psi \left(\frac{b}{a}\right)}{a}+\frac{a+2 b}{2 a^2 p}+O\left(\frac{1}{p^2}\right)$$ $$S_p(a,b)-S_p(a,b+1)=\frac{\psi \left(\frac{b+1}{a}\right)-\psi \left(\frac{b}{a}\right)}{a}-\frac{1}{a^2 p}+O\left(\frac{1}{p^2}\right)$$ In your case $b \ll a$ and $$\psi(\epsilon)=-\frac{1}{\epsilon }-\gamma +\frac{\pi ^2 }{6}\epsilon+O\left(\epsilon ^2\right)$$ Using it twice $$\frac{\psi \left(\frac{b+1}{a}\right)-\psi \left(\frac{b}{a}\right)}{a}=\frac{\pi ^2}{6 a^2}+\frac{1}{b^2+b}+\cdots$$ $$S_p(a,b)-S_p(a,b+1)\sim\frac{\pi ^2}{6 a^2}+\frac{1}{b^2+b}-\frac{1}{a^2 p}+O\left(\frac{1}{p^2}\right)$$ So, with your numbers, the result is $$S_\infty(22,1)-S_\infty(22,2)) \sim \frac{1}{2}+\frac{\pi ^2}{2904}=0.5034$$ while the exact value is $0.5031$.

Adding one more term in the last expansion, the result would have been $$S_\infty(22,1)-S_\infty(22,2)) \sim \frac{1}{2}+\frac{\pi ^2}{2904}-\frac{3 \zeta (3)}{10648}=0.5030600$$ while the exact value is $0.5030900$.

Just by curiosity, have a look at the exact result given by Wolfram Alpha.

Claude Leibovici
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