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Question Number 128048 by bramlexs22 last updated on 04/Jan/21

$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{8\mathrm{n}+3}=?$$

Answered by Olaf last updated on 04/Jan/21

$$\mathrm{L}(\lambda ,\alpha ,s)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{e^{2i\pi \lambda n}}{{(n+\alpha )}^s}\mathrm{or}\mathrm{Lerch}(\lambda ,\alpha ,s)$$$$\left(\mathrm{zeta}\mathrm{Lerch}\mathrm{function}\right)$$$$\mathrm{Let}\mathrm{S}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{8n+3}$$$$\mathrm{S}=\frac{1}{8}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{e^{i\pi n}}{{(n+\frac{3}{8})}^1}=\frac{1}{8}\mathrm{L}(\frac{1}{2},\frac{3}{8},1)$$$$\mathrm{We}\mathrm{have}\mathrm{too}:$$$$\mathrm{\Phi }(z,s,\alpha )=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{z^n}{{(n+\alpha )}^s}\mathrm{or}\mathrm{LerchPhi}(z,s,\alpha )$$$$\left(\mathrm{transcendent}\mathrm{Lerch}\mathrm{function}\right)$$$$\mathrm{S}=\frac{1}{8}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(n+\frac{3}{8})}^1}=\frac{1}{8}\mathrm{\Phi }(-1,1,\frac{3}{8})$$$$\mathrm{S}\approx 0,2738982192$$

Commented by bramlexs22 last updated on 04/Jan/21

I just found out about the lerch zeta function

Commented by bramlexs22 last updated on 04/Jan/21

$$\mathrm{how}\mathrm{about}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{8\mathrm{n}+5}\mathrm{sir}?$$

Commented by Olaf last updated on 04/Jan/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{8n+5}=\frac{1}{8}\mathrm{LerchPhi}(-1,1,\frac{5}{8})$$$$\approx 0,1511562039$$

Commented by bramlexs22 last updated on 04/Jan/21

$$\mathrm{sir}\mathrm{how}\mathrm{you}\mathrm{get}\approx 0.1511562039?$$$$\mathrm{by}\mathrm{software}\mathrm{sir}\mathrm{or}\mathrm{calculator}?$$

Answered by mindispower last updated on 04/Jan/21

$$=\mathrm{\Sigma }\frac{1}{16(n+\frac{3}{16})}-\frac{1}{16(n+\frac{11}{16})}$$$$=\mathrm{\Sigma }\frac{2}{256(n+\frac{3}{16})(n+\frac{11}{16})}$$$$=\frac{1}{16}(\mathrm{\Psi }\left(\frac{3}{16}\right)-\mathrm{\Psi }\left(\frac{11}{16}\right))$$$$4lose$$$$2ndway{\int }_0^1{x}^2{(-x^8)}^{n}dx=\frac{{(-1)}^n}{8n+1}dx$$$$justusingbasicresults\mathrm{\Sigma }{(-z)}^n=\frac{1}{1+z}$$$$and{Z}^n+1=0\Rightarrow Z=e^{i\left(\frac{2k+1}{n}\right)\pi }$$$$\Rightarrow \sum _{n⩾0}{\int }_0^1{x}^2.{(-x^8)}^{n}dx=\sum _{n⩾0}\frac{{(-1)}^n}{8n+3}$$$$\iff {\int }_0^1\frac{x^2}{1+x^8}dx=\sum _{n⩾0}\frac{{(-1)}^n}{8n+3}$$$$x^8+1=\underset{k=0}{\prod ^7}(X-e^{i\left(\frac{(2k+1)\pi }{8}\right))})$$$$=\underset{k=0}{\prod ^3}(X^2-2cos\left(\frac{2k+1}{8}\pi \right)X+1)$$$$decompositionwegetelementry\int \frac{dx}{X^2+aX+1}$$$$findvalueofintegrals$$

Answered by Dwaipayan Shikari last updated on 04/Jan/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{8n+3}={\int }_0^1\sum ^{\mathrm{\infty }}{(-1)}^n{x}^{8n+2}dx$$$$={\int }_0^1\frac{x^2}{1+x^8}dxwecansolvethisAfterdecomposing$$$$$$

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