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Question Number 41409 by maxmathsup by imad last updated on 06/Aug/18

calculate S_p =Σ_(n=1) ^∞ (((−1)^n )/(n(n+1)(n+2)...(n+p))) with p fromN

$${calculate}\:{S}_{{p}} =\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left({n}+{p}\right)}\:\:{with}\:{p}\:{fromN} \\ $$

Answered by sma3l2996 last updated on 07/Aug/18

(1/(n(n+1)(n+2)...(n+p)))=(a_0 /n)+(a_1 /(n+1))+...+(a_p /(n+p)) with a_i =lim_(n→−i) ((n+i)/(n(n+1)...(n+p))) (1/(n(n+1)(n+2)...(n+p)))=Σ_(i=0) ^p (a_i /(n+i)) so S_p =Σ_(n=1) ^∞ (((−1)^n )/(n(n+1)(n+2)...(n+p)))=Σ_(n=1) ^∞ Σ_(i=0) ^p (((−1)^n a_i )/(n+i)) =Σ_(i=0) ^p a_i Σ_(n=1) ^∞ (((−1)^n )/(n+i)) let m=n+i S_n =Σ_(i=0) ^p a_i Σ_(m=i+1) ^∞ (((−1)^(m−i) )/m) =Σ_(i=0) ^∞ a_i (−1)^(−i) (Σ_(m=i+1) ^∞ (((−1)^m )/m)+Σ_(m=1) ^i (((−1)^m )/m)−Σ_(m=1) ^i (((−1)^m )/m)) =Σ_(i=0) ^p (−1)^i a_i (Σ_(m=1) ^∞ (((−1)^m )/m)−Σ_(m=1) ^i (((−1)^m )/m)) note: Σ_(n=1) ^∞ (((−1)^n )/n)=−ln(2) Therefore: S_p =Σ_(i=0) ^p (−1)^i a_i (−ln(2)−Σ_(m=1) ^i (((−1)^m )/m))

$$\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left({n}+{p}\right)}=\frac{{a}_{\mathrm{0}} }{{n}}+\frac{{a}_{\mathrm{1}} }{{n}+\mathrm{1}}+...+\frac{{a}_{{p}} }{{n}+{p}} \\ $$$${with}\:\:{a}_{{i}} =\underset{{n}\rightarrow−{i}} {{lim}}\frac{{n}+{i}}{{n}\left({n}+\mathrm{1}\right)...\left({n}+{p}\right)} \\ $$$$\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left({n}+{p}\right)}=\underset{{i}=\mathrm{0}} {\overset{{p}} {\sum}}\frac{{a}_{{i}} }{{n}+{i}} \\ $$$${so}\:\:\:{S}_{{p}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left({n}+{p}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{{p}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {a}_{{i}} }{{n}+{i}} \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{{p}} {\sum}}{a}_{{i}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+{i}} \\ $$$${let}\:\:\:{m}={n}+{i} \\ $$$${S}_{{n}} =\underset{{i}=\mathrm{0}} {\overset{{p}} {\sum}}{a}_{{i}} \underset{{m}={i}+\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}−{i}} }{{m}} \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} \left(−\mathrm{1}\right)^{−{i}} \left(\underset{{m}={i}+\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}}+\underset{{m}=\mathrm{1}} {\overset{{i}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}}−\underset{{m}=\mathrm{1}} {\overset{{i}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}}\right) \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{{p}} {\sum}}\left(−\mathrm{1}\right)^{{i}} {a}_{{i}} \left(\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}}−\underset{{m}=\mathrm{1}} {\overset{{i}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}}\right) \\ $$$${note}:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}=−{ln}\left(\mathrm{2}\right) \\ $$$${Therefore}:\:\: \\ $$$${S}_{{p}} =\underset{{i}=\mathrm{0}} {\overset{{p}} {\sum}}\left(−\mathrm{1}\right)^{{i}} {a}_{{i}} \left(−{ln}\left(\mathrm{2}\right)−\underset{{m}=\mathrm{1}} {\overset{{i}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}}\right) \\ $$

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