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Question Number 126200 by Dwaipayan Shikari last updated on 18/Dec/20

$$\frac{1}{1^2}-\frac{1}{2^3}+\frac{1}{3^5}-\frac{1}{4^7}+\frac{1}{5^{11}}-\frac{1}{6^{13}}+....$$

Commented by Olaf last updated on 18/Dec/20

$$$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{n^{p_n}}$$$$p_n=\mathrm{Nth}\mathrm{prime}\mathrm{number}$$$$\mathrm{Is}\mathrm{it}\mathrm{the}\mathrm{question}\mathrm{sir}?$$

Commented by Dwaipayan Shikari last updated on 18/Dec/20

$$yessir!$$

Commented by Olaf last updated on 18/Dec/20

$$$$$$\mathrm{I}\mathrm{have}\mathrm{no}\mathrm{idea}\mathrm{how}\mathrm{to}\mathrm{calculate}\mathrm{this}\mathrm{but}$$$$\mathrm{maybe}\mathrm{we}\mathrm{can}\mathrm{find}\mathrm{lower}\mathrm{and}\mathrm{upper}$$$$\mathrm{bounds}:$$$$p_n\mathrm{Nth}\mathrm{prime}\mathrm{number}$$$$1,2,3,5,7,11,13...$$$$$$$$\mathrm{Felgner}\mathrm{result}\mathrm{in}1990:$$$$0,91n\ln n<p_n<1,7n\ln n$$$$\mathrm{S}=\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(n-1)}^{p_n}}$$$$\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(n-1)}^{1,7n\ln n}}<\mathrm{S}<\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(n-1)}^{0,91n\ln n}}$$$$...$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{if}\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{good}\mathrm{way}.$$

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