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$$\sum_{n=0}^\infty\left(\frac1{50+101n}-\frac1{51+101n}\right)$$

How to prove that the value of the above summation is equal to $(\pi/101)\tan(\pi/202)?$ I am trying this question by putting n=0,1,2,3,...and so on and getting the series to be $$(1/50)-(1/51)+(1/151)-(1/152)+(1/252)-(1/253)+...$$and so on but how to prove that this summation will be equal to $(\pi/101)\tan(\pi/202)?$ Please help me out.

Coherent Sheaf
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SUBHA SANKAR ROY
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2 Answers2

3

Too long for a comment

First, we note that $$S_a=\sum_{n=0}^\infty\left(\frac1{a+(2a+1)n}-\frac1{a+1+(2a+1)n}\right)$$ $$=\sum_{n=0}^\infty\frac1{(a+(2a+1)n)(a+1+(2a+1)n)}$$ $$=\frac12\sum_{n=-\infty}^\infty\frac1{(a+(2a+1)n)(a+1+(2a+1)n)}=\frac1{2(2a+1)^2}\sum_{n=-\infty}^\infty\frac1{(n-\frac a{2a+1})(n-\frac {a+1}{2a+1})}$$ Using a standart approach and integrating along a big circle ($R\to\infty$) in the complex plane $$\oint\frac{\pi\cot\pi z}{(z-\frac a{2a+1})(z-\frac {a+1}{2a+1})}dz=0=2\pi i\sum\operatorname{Res}$$ $$\Rightarrow\, \sum_{n=-\infty}^\infty\frac1{(n-\frac a{2a+1})(n-\frac {a+1}{2a+1})}+\frac{\pi\cot(\pi\frac a{2a+1})}{\frac a{2a+1}-\frac {a+1}{2a+1}}+\frac{\pi\cot(\pi\frac {a+1}{2a+1})}{\frac {a+1}{2a+1}-\frac a{2a+1}}=0$$ $$\Rightarrow\,\sum_{n=-\infty}^\infty\frac1{(n-\frac a{2a+1})(n-\frac {a+1}{2a+1})}=-(2a+1)\pi\left(\cot(\pi\frac {a+1}{2a+1})-\cot(\pi\frac a{2a+1})\right)$$ $$=-(2a+1)\pi\left(\cot(\pi\frac {a+1+a-a}{2a+1})-\cot(\pi\frac a{2a+1})\right)$$ $$=2\pi(2a+1)\cot\frac{\pi a}{2a+1}=2\pi(2a+1)\tan\left(\frac\pi2-\frac{\pi a}{2a+1}\right)=2\pi(2a+1)\tan\frac{\pi }{4a+2}$$ $$\Rightarrow\, S_a=\frac\pi{2a+1}\tan\frac\pi{4a+2}$$

Svyatoslav
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1

Consider the more general case of the partial sum $$S_p=\sum_{n=0}^p\left(\frac1{a+b\, n}-\frac1{(a+1)+b\, n}\right)$$ $$S_p=\frac 1 b \left(\psi \left(\frac{a+b}{b}+p\right)-\psi \left(\frac{a+b+1}{b}+p\right)-\psi \left(\frac{a}{b}\right)+\psi \left(\frac{a+1}{b}\right)\right)$$

Expanding as a series for large values of $p$ $$S_p=\frac 1 b \left(\psi \left(\frac{a+1}{b}\right)-\psi \left(\frac{a}{b}\right)\right)-\frac{1}{b^2 p}+O\left(\frac{1}{p^2}\right)$$ For the particular case where $b=2a+1$ $$S_p=\frac 1 {2a+1} \left(\psi \left(\frac{a+1}{2 a+1}\right)-\psi \left(\frac{a}{2 a+1}\right)\right)-\frac 1 {(2a+1)^2\, p}+O\left(\frac{1}{p^2}\right)$$ The first term write $$\frac \pi {2a+1} \tan \left(\frac{\pi }{2(2 a+1)}\right)$$ because of the identity

$$\psi (1-z)-\psi (z)=\pi \cot (\pi z)$$

Claude Leibovici
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