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Question Number 72912 by mathmax by abdo last updated on 04/Nov/19

$$calculate{S}_p=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+p}$$

Answered by mind is power last updated on 04/Nov/19

$${\mathrm{S}}_{\mathrm{p}}=\sum _{\mathrm{n}⩾\mathrm{p}}\frac{{(-1)}^{\mathrm{k}-\mathrm{p}}}{\mathrm{k}}$$$$\mathrm{p}⩾2$$$$={(-1)}^{\mathrm{p}}.\sum _{\mathrm{n}⩾\mathrm{p}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}={(-1)}^{\mathrm{p}}.(\sum _{\mathrm{n}⩾1}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}-\underset{\mathrm{n}=1}{\sum ^{\mathrm{p}-1}}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}})$$$$={(-1)}^{\mathrm{p}}.(-\ln \left(2\right)-\underset{\mathrm{n}=1}{\sum ^{\mathrm{p}-1}}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}})$$

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