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How to evaluate the Nth sum of the series below? $$ \sum_{n=0}^{N} (\frac{1}{4n+1}) - (\frac{1}{4n+3})$$ I have tried algebraic manipulation, geometric series formula, and the difference method but nothing works. What could be the method?

JMP
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    I do not think there is a nice answer. The simplest form you can get involves digamma functions: $$ \tfrac{1}{4}\left( {\psi \left( {N + \tfrac{5}{4}} \right) - \psi \left( {N+ \tfrac{5}{4} + \tfrac{1}{2}} \right) + \pi } \right). $$ I am not aware of any simple formula for ${\psi (z) - \psi \left( {z + \frac{1}{2}} \right)}$. – Gary May 01 '21 at 10:13
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    Just note that to evaluate the infinite sum in the title, you do not necessarily need a formula for n-th partial sum (if you recognize a well-known series $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots$, which is just a series for $\arctan x$ with $x=1$ ) – Sil May 01 '21 at 11:06
  • @Sil What is the proof? I really want to show that the infinite sum is the inverse of tan. –  May 01 '21 at 11:14
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    You can see for example Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$? and linked posts – Sil May 01 '21 at 11:17

2 Answers2

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Write $\displaystyle \sum_{n=1}^N\left(\dfrac{1}{4n+1} - \dfrac{1}{4n+3}\right) = \displaystyle \int_{0}^{1} \displaystyle \sum_{n=1}^N(x^{4n}-x^{4n+2})dx= \displaystyle \int_{0}^1\displaystyle \sum_{n=1}^N x^{4n}dx-\displaystyle \int_{0}^1\displaystyle \sum_{n=1}^Nx^{4n+2}dx$. Note that both integrands are finite geometric series. You can find the sum. After that perhaps you would have to use the more advanced functions to deal with it since there might not be an elementary anti derivatives.

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Similar to @Gary's answer but using the generalized harmonic numbers $$S_N=\sum_{n=0}^N\left(\dfrac{1}{4n+1} - \dfrac{1}{4n+3}\right)=\frac{1}{4} \left(H_{N+\frac{1}{4}}-H_{N+\frac{3}{4}}+\pi \right)$$ Now, if $N$ is large, using the asymptotics $$S_N=\frac{\pi }{4}-\frac{1}{8 N}+\frac{1}{8 N^2}-\frac{15}{128 N^3}+\frac{13}{128 N^4}+O\left(\frac{1}{N^5}\right)$$

In practice, the relative error is smaller than $0.1$% as soon as $N \geq 3$ and smaller than $0.01$% as soon as $N \geq 4$.

Claude Leibovici
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