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

calculate S_p = Σ_(n=0) ^∞ (((−1)^n )/(n+p))

$${calculate}\:{S}_{{p}} =\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+{p}} \\ $$

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

S_p =Σ_(n≥p) (((−1)^(k−p) )/k) p≥2 =(−1)^p .Σ_(n≥p) (((−1)^k )/k)=(−1)^p .(Σ_(n≥1) (((−1)^n )/n)−Σ_(n=1) ^(p−1) (((−1)^n )/n)) =(−1)^p .(−ln(2)−Σ_(n=1) ^(p−1) (((−1)^n )/n))

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

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