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Question Number 66483 by ~ À ® @ 237 ~ last updated on 16/Aug/19

calculate Σ_(k=2) ^∞ (((−1)^k )/k) ζ(k) if ζ(s)=Σ_(n=1) ^∞ (1/n^s )

$$\:{calculate}\:\:\:\:\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:\zeta\left({k}\right)\:\:\:\:\:\:\:{if}\:\:\:\:\zeta\left({s}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{1}}{{n}^{{s}} }\: \\ $$

Commented by mathmax by abdo last updated on 16/Aug/19

let S =Σ_(k=2) ^∞ (((−1)^k )/k)ξ(k) ⇒ S =Σ_(k=2) ^∞ (((−1)^k )/k)(Σ_(n=1) ^∞ (1/n^k )) =Σ_(n=1) ^∞ (Σ_(k=2) ^∞ (((−1)^k )/(k n^k ))) =Σ_(n=1) ^∞ (Σ_(k=2) ^∞ (1/k)(−(1/n))^k ) let find f(x) =Σ_(k=2) ^∞ (x^k /k) with ∣x∣<1⇒f^′ (x)=Σ_(k=2) ^∞ x^(k−1) =Σ_(k=1) ^∞ x^k =x Σ_(k=1) ^∞ x^(k−1) [=x Σ_(k=0) ^∞ x^k =(x/(1−x)) ⇒f(x) =∫(x/(1−x))dx +c =−∫(x/(x−1))dx+c =−∫((x−1+1)/(x−1))dx +c =−x+ln∣x−1∣ +c f(0) =0=c ⇒f(x) =−x+ln∣x−1∣⇒ Σ_(k=2) ^∞ (1/k)(−(1/n))^k =(1/n)+ln∣(1/n)+1∣ =(1/n) +ln(((n+1)/n)) =(1/n)+ln(n)−ln(n+1) ⇒ S =Σ_(n=1) ^∞ ((1/n) +ln(n)−ln(n+1))=lim_(n→+∞) S_n with S_n =Σ_(k=1) ^n ((1/k)+ln(k)−ln(k+1)) =Σ_(k=1) ^n (1/k) +Σ_(k=1) ^n (ln(k)−ln(k+1)) =H_n −ln(n+1) =H_n −ln(n)−ln(1+(1/n))→γ (n→+∞) ⇒ Σ_(k=2) ^∞ (((−1)^k )/k)ξ(k) =γ (constant of Euler)

$${let}\:{S}\:=\sum_{{k}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\xi\left({k}\right)\:\Rightarrow\:{S}\:=\sum_{{k}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{{k}} }\right) \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(\sum_{{k}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}\:{n}^{{k}} }\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \left(\sum_{{k}=\mathrm{2}} ^{\infty} \frac{\mathrm{1}}{{k}}\left(−\frac{\mathrm{1}}{{n}}\right)^{{k}} \right) \\ $$$${let}\:{find}\:{f}\left({x}\right)\:=\sum_{{k}=\mathrm{2}} ^{\infty} \frac{{x}^{{k}} }{{k}}\:{with}\:\mid{x}\mid<\mathrm{1}\Rightarrow{f}^{'} \left({x}\right)=\sum_{{k}=\mathrm{2}} ^{\infty} {x}^{{k}−\mathrm{1}} \:=\sum_{{k}=\mathrm{1}} ^{\infty} {x}^{{k}} \\ $$$$={x}\:\sum_{{k}=\mathrm{1}} ^{\infty} \:{x}^{{k}−\mathrm{1}} \left[={x}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:{x}^{{k}} \:=\frac{{x}}{\mathrm{1}−{x}}\:\Rightarrow{f}\left({x}\right)\:=\int\frac{{x}}{\mathrm{1}−{x}}{dx}\:+{c}\right. \\ $$$$=−\int\frac{{x}}{{x}−\mathrm{1}}{dx}+{c}\:=−\int\frac{{x}−\mathrm{1}+\mathrm{1}}{{x}−\mathrm{1}}{dx}\:+{c}\:=−{x}+{ln}\mid{x}−\mathrm{1}\mid\:+{c} \\ $$$${f}\left(\mathrm{0}\right)\:=\mathrm{0}={c}\:\Rightarrow{f}\left({x}\right)\:=−{x}+{ln}\mid{x}−\mathrm{1}\mid\Rightarrow \\ $$$$\sum_{{k}=\mathrm{2}} ^{\infty} \frac{\mathrm{1}}{{k}}\left(−\frac{\mathrm{1}}{{n}}\right)^{{k}} \:=\frac{\mathrm{1}}{{n}}+{ln}\mid\frac{\mathrm{1}}{{n}}+\mathrm{1}\mid\:=\frac{\mathrm{1}}{{n}}\:+{ln}\left(\frac{{n}+\mathrm{1}}{{n}}\right) \\ $$$$=\frac{\mathrm{1}}{{n}}+{ln}\left({n}\right)−{ln}\left({n}+\mathrm{1}\right)\:\Rightarrow \\ $$$${S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(\frac{\mathrm{1}}{{n}}\:+{ln}\left({n}\right)−{ln}\left({n}+\mathrm{1}\right)\right)={lim}_{{n}\rightarrow+\infty} \:\:{S}_{{n}} \:{with} \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \left(\frac{\mathrm{1}}{{k}}+{ln}\left({k}\right)−{ln}\left({k}+\mathrm{1}\right)\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}}\:+\sum_{{k}=\mathrm{1}} ^{{n}} \left({ln}\left({k}\right)−{ln}\left({k}+\mathrm{1}\right)\right) \\ $$$$={H}_{{n}} −{ln}\left({n}+\mathrm{1}\right)\:={H}_{{n}} −{ln}\left({n}\right)−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\rightarrow\gamma\:\left({n}\rightarrow+\infty\right)\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\xi\left({k}\right)\:=\gamma\:\:\:\:\:\:\:\left({constant}\:{of}\:{Euler}\right) \\ $$

Commented by mathmax by abdo last updated on 17/Aug/19

f(x)=−x−ln∣x−1∣⇒Σ_(k=2) ^∞ (1/k)(−(1/n))^k =(1/n)−ln((1/n)+1) =(1/n) −ln(((n+1)/n)) =(1/n)+ln(n)−ln(n+1)...(error of typo)

$${f}\left({x}\right)=−{x}−{ln}\mid{x}−\mathrm{1}\mid\Rightarrow\sum_{{k}=\mathrm{2}} ^{\infty} \frac{\mathrm{1}}{{k}}\left(−\frac{\mathrm{1}}{{n}}\right)^{{k}} \:=\frac{\mathrm{1}}{{n}}−{ln}\left(\frac{\mathrm{1}}{{n}}+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{{n}}\:−{ln}\left(\frac{{n}+\mathrm{1}}{{n}}\right)\:=\frac{\mathrm{1}}{{n}}+{ln}\left({n}\right)−{ln}\left({n}+\mathrm{1}\right)...\left({error}\:{of}\:{typo}\right) \\ $$

Commented by ~ À ® @ 237 ~ last updated on 17/Aug/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by mathmax by abdo last updated on 17/Aug/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

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