Calculus-n-1-1-n-k-1-n-k-2-

Question Number 143603 by mnjuly1970 last updated on 16/Jun/21

.....Calculus..... Ω:=Σ_(n=1) ^∞ (1/(n^k (1+n))) (k≥ 2) ......

$$\:\:\:\:\:\:\:\:\:\:\:\:\:…..{Calculus}….. \\ $$$$\:\:\:\:\:\:\:\:\Omega:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{k}} \left(\mathrm{1}+{n}\right)}\:\:\:\left({k}\geqslant\:\mathrm{2}\right)\:…… \\ $$

Answered by Dwaipayan Shikari last updated on 16/Jun/21

Ω:=Σ_(n=1) ^∞ (1/(n^k (n+1)))=∫_0 ^1 Li_k (x)dx =[xLi_k (x)]_0 ^1 −∫_0 ^1 Σ_(n=1) ^∞ n ((x^(n−1) .x)/n^k )dx =Li_k (1)−∫_0 ^1 Σ_(n=1) ^∞ (x^n /n^(k−1) )dx=Li_k (1)−∫_0 ^1 Li_(k−1) (1)dx =Li_k (1)−Li_(k−1) (1)+Li_(k−2) (1)−...Li_2 (1) =ζ(k)−ζ(k−1)+ζ(k−2)−ζ(k−3)+...+(π^2 /6) if k odd or ζ(k)−ζ(k−1)+...−(π^2 /6) if k even

$$\Omega:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{k}} \left({n}+\mathrm{1}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} {Li}_{{k}} \left({x}\right){dx} \\ $$$$=\left[{xLi}_{{k}} \left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}\:\frac{{x}^{{n}−\mathrm{1}} .{x}}{{n}^{{k}} }{dx} \\ $$$$={Li}_{{k}} \left(\mathrm{1}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{{k}−\mathrm{1}} }{dx}={Li}_{{k}} \left(\mathrm{1}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} {Li}_{{k}−\mathrm{1}} \left(\mathrm{1}\right){dx} \\ $$$$={Li}_{{k}} \left(\mathrm{1}\right)−{Li}_{{k}−\mathrm{1}} \left(\mathrm{1}\right)+{Li}_{{k}−\mathrm{2}} \left(\mathrm{1}\right)−…{Li}_{\mathrm{2}} \left(\mathrm{1}\right) \\ $$$$=\zeta\left({k}\right)−\zeta\left({k}−\mathrm{1}\right)+\zeta\left({k}−\mathrm{2}\right)−\zeta\left({k}−\mathrm{3}\right)+…+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:{if}\:{k}\:{odd} \\ $$$${or}\:\zeta\left({k}\right)−\zeta\left({k}−\mathrm{1}\right)+…−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:{if}\:{k}\:{even} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 16/Jun/21

$$\:\:\:\:\:{thank}\:{you}\:{so}\:{much}… \\ $$

Answered by mindispower last updated on 16/Jun/21

f(k)=Σ_(n≥1) (1/(n^k (n+1))),f(1)=1 f(k)+f(k−1)=Σ(1/n^k )=ζ(k),k≥1 (−1)^m f(m)+(−1)^m f(m−1)=(−1)^m ζ(m) Σ_(m=2) ^k (−1)^m f(m)+(−1)^m f(m−1)=Σ_(m=2) ^k (−1)^m ζ(m) (−1)^k f(k)+f(1)=Σ_(m=2) ^k (−1)^m ζ(m) f(k)=(−1)^k Σ_(m=2) ^k (−1)^m ζ(m)+(−1)^(k−1) ,k≥2 f(1)=1

$${f}\left({k}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{{k}} \left({n}+\mathrm{1}\right)},{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${f}\left({k}\right)+{f}\left({k}−\mathrm{1}\right)=\Sigma\frac{\mathrm{1}}{{n}^{{k}} }=\zeta\left({k}\right),{k}\geqslant\mathrm{1} \\ $$$$\left(−\mathrm{1}\right)^{{m}} {f}\left({m}\right)+\left(−\mathrm{1}\right)^{{m}} {f}\left({m}−\mathrm{1}\right)=\left(−\mathrm{1}\right)^{{m}} \zeta\left({m}\right) \\ $$$$\underset{{m}=\mathrm{2}} {\overset{{k}} {\sum}}\left(−\mathrm{1}\right)^{{m}} {f}\left({m}\right)+\left(−\mathrm{1}\right)^{{m}} {f}\left({m}−\mathrm{1}\right)=\underset{{m}=\mathrm{2}} {\overset{{k}} {\sum}}\left(−\mathrm{1}\right)^{{m}} \zeta\left({m}\right) \\ $$$$\left(−\mathrm{1}\right)^{{k}} {f}\left({k}\right)+{f}\left(\mathrm{1}\right)=\underset{{m}=\mathrm{2}} {\overset{{k}} {\sum}}\left(−\mathrm{1}\right)^{{m}} \zeta\left({m}\right) \\ $$$${f}\left({k}\right)=\left(−\mathrm{1}\right)^{{k}} \underset{{m}=\mathrm{2}} {\overset{{k}} {\sum}}\left(−\mathrm{1}\right)^{{m}} \zeta\left({m}\right)+\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} ,{k}\geqslant\mathrm{2} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$

Commented by mnjuly1970 last updated on 16/Jun/21

$$\:{very}\:{nice}\:…. \\ $$

Commented by mindispower last updated on 16/Jun/21

$${pleasur} \\ $$

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