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Question Number 143045 by 0731619 last updated on 09/Jun/21

Answered by Dwaipayan Shikari last updated on 09/Jun/21

$${\int }_0^1\frac{1-x^n}{1-x}dx=-\gamma +\psi (n+1)$$$$\underset{n\to 3}{\lim }\frac{-\gamma -1.833+\psi (n+1)}{\mathrm{\Gamma }(n+1)-6}\stackrel{Lhopital}{=}\frac{{\psi }^{\prime }(n+1)}{{\mathrm{\Gamma }}^{\prime }(n+1)}=\frac{{\psi }^{\prime }\left(4\right)}{\psi \left(4\right)\mathrm{\Gamma }\left(4\right)}$$$$\frac{\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+4)}^2}}{(\gamma +1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})6}=\frac{\frac{{\pi }^2}{6}-\frac{1}{3^2}-\frac{1}{2^2}-1}{6\gamma +\frac{25}{2}}=\frac{6{\pi }^2-294}{216\gamma +450}$$

Commented by Canebulok last updated on 11/Jun/21

Hello mr. Payan , are you a mathematician?

Commented by Dwaipayan Shikari last updated on 11/Jun/21

Thanks for asking. I am a student and I am trying to learn my interest

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