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Question Number 138065 by mnjuly1970 last updated on 09/Apr/21
                   ...nice ... ... ... calculus...         prove::           Ω=Σ_(n=1) ^∞ ((ζ(2n+1)−1)/(n+1)) =−γ+log(2)
$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:…{nice}\:…\:…\:…\:{calculus}… \\ $$$$\:\:\:\:\:\:\:{prove}:: \\ $$$$\:\:\:\:\:\:\:\:\:\Omega=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\zeta\left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{1}}{{n}+\mathrm{1}}\:=−\gamma+{log}\left(\mathrm{2}\right) \\ $$
Answered by Ñï= last updated on 09/Apr/21
Ω=Σ_(k=2) ^∞ Σ_(n=1) ^∞ (1/((n+1)k^(2n+1) ))=−Σ_(k=2) ^∞ ((1/k)+ln(1−(1/k^2 )))  =lim_(m→∞) (−H_m +1−Σ_(k=2) ^∞ (kln(k+1)+kln(k−1)−2klnk))  =lim_(m→∞) (−H_m +1+ln2+lnm−mln(1+(1/m)))  =lim_(m→∞) (−H_m +lnm+ln2+O((1/m)))  =−γ+ln2  ...... i dont understand...just copy...
$$\Omega=\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){k}^{\mathrm{2}{n}+\mathrm{1}} }=−\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{k}}+{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)\right) \\ $$$$=\underset{{m}\rightarrow\infty} {\mathrm{lim}}\left(−{H}_{{m}} +\mathrm{1}−\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\left({kln}\left({k}+\mathrm{1}\right)+{kln}\left({k}−\mathrm{1}\right)−\mathrm{2}{klnk}\right)\right) \\ $$$$=\underset{{m}\rightarrow\infty} {\mathrm{lim}}\left(−{H}_{{m}} +\mathrm{1}+{ln}\mathrm{2}+{lnm}−{mln}\left(\mathrm{1}+\frac{\mathrm{1}}{{m}}\right)\right) \\ $$$$=\underset{{m}\rightarrow\infty} {\mathrm{lim}}\left(−{H}_{{m}} +{lnm}+{ln}\mathrm{2}+\mathcal{O}\left(\frac{\mathrm{1}}{{m}}\right)\right) \\ $$$$=−\gamma+{ln}\mathrm{2} \\ $$$$……\:{i}\:{dont}\:{understand}…{just}\:{copy}… \\ $$
Commented by mnjuly1970 last updated on 10/Apr/21
Commented by mnjuly1970 last updated on 10/Apr/21

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