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Question Number 221387 by Nicholas666 last updated on 02/Jun/25
Σ_(k = 1) ^∞ (2Σ_(n = 1) ^∞ (1/(n^2 + kn)))^2 = ?
$$ \\ $$$$\:\:\:\:\:\:\:\underset{{k}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{2}\underset{{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\:{kn}}\right)^{\mathrm{2}} \:=\:? \\ $$$$ \\ $$
Answered by MrGaster last updated on 05/Jun/25
Σ_(k = 1) ^∞ (2Σ_(n = 1) ^∞ (1/(n^2 + kn)))^2 =4Σ_(k=1) ^∞ (Σ_(n=1) ^∞ (1/(n(n+k))))^2 =4Σ_(n=1) ^∞ ((H_k /k))^2 =4Σ_(k=1) ^∞ (H_k ^2 /k^2 ) =4∙((17π^4 )/(360)) =((17π^4 )/(90)) (2):=4Σ_(k=1) ^∞ ((H_k /k))^2 =4Σ_(k=1) ^∞ (H_k ^2 /k^2 ) =4∙((17π^4 )/(360)) =((17π^4 )/(90)) (3)Σ_(n=1) ^∞ (1/(n(n+k))) =(1/k)Σ_(n=1) ^∞ ((1/n)−(1/(n−k))) =(H_k /k) ⇒2Σ_(n=1) ^∞ (1/(n^2 +kn))=((2H_k )/k) ∴ Σ_(k = 1) ^∞ (2Σ_(n = 1) ^∞ (1/(n^2 + kn)))^2 =Σ_(k=1) ^∞ (((2H_k )/k))^2 =4Σ_(k=1) ^∞ (H_k ^2 /k^2 ) =4∙((17)/4)ζ(4) =17ζ(4) =17∙(π^4 /(90)) =((17π^4 )/(90))
$$\:\:\:\:\underset{{k}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{2}\underset{{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\:{kn}}\right)^{\mathrm{2}} \:=\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+{k}\right)}\right)^{\mathrm{2}} \\ $$$$=\mathrm{4}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{H}_{{k}} }{{k}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{k}} ^{\mathrm{2}} }{{k}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\centerdot\frac{\mathrm{17}\pi^{\mathrm{4}} }{\mathrm{360}} \\ $$$$=\frac{\mathrm{17}\pi^{\mathrm{4}} }{\mathrm{90}} \\ $$$$\left(\mathrm{2}\right):=\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{H}_{{k}} }{{k}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{k}} ^{\mathrm{2}} }{{k}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\centerdot\frac{\mathrm{17}\pi^{\mathrm{4}} }{\mathrm{360}} \\ $$$$=\frac{\mathrm{17}\pi^{\mathrm{4}} }{\mathrm{90}} \\ $$$$\left(\mathrm{3}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+{k}\right)} \\ $$$$=\frac{\mathrm{1}}{{k}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}−{k}}\right) \\ $$$$=\frac{{H}_{{k}} }{{k}} \\ $$$$\Rightarrow\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{kn}}=\frac{\mathrm{2}{H}_{{k}} }{{k}} \\ $$$$\therefore\:\underset{{k}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{2}\underset{{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\:{kn}}\right)^{\mathrm{2}} \:=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{2}{H}_{{k}} }{{k}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{k}} ^{\mathrm{2}} }{{k}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\centerdot\frac{\mathrm{17}}{\mathrm{4}}\zeta\left(\mathrm{4}\right) \\ $$$$=\mathrm{17}\zeta\left(\mathrm{4}\right) \\ $$$$=\mathrm{17}\centerdot\frac{\pi^{\mathrm{4}} }{\mathrm{90}} \\ $$$$=\frac{\mathrm{17}\pi^{\mathrm{4}} }{\mathrm{90}} \\ $$

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