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n-1-n-n-2-log-n-1-In-explicit-form-




Question Number 124718 by Dwaipayan Shikari last updated on 05/Dec/20
Σ_(n=1) ^n n^2 log(n+1)    (In explicit form )
$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{n}^{\mathrm{2}} {log}\left({n}+\mathrm{1}\right)\:\:\:\:\left({In}\:{explicit}\:{form}\:\right) \\ $$
Commented by Dwaipayan Shikari last updated on 05/Dec/20
In Ramanujan′s Note book  1^2 log(1)+2^2 log(2)+3^2 log(3)+4^2 log(4)+...+n^2 log(n)  =((n(n+1)(2n+1))/6)log(n)−(n^3 /9)+((ζ(3))/(4π^2 ))+(n/(12))−(1/(360n))+...+C    But i haven′t found any proof
$${In}\:{Ramanujan}'{s}\:{Note}\:{book} \\ $$$$\mathrm{1}^{\mathrm{2}} {log}\left(\mathrm{1}\right)+\mathrm{2}^{\mathrm{2}} {log}\left(\mathrm{2}\right)+\mathrm{3}^{\mathrm{2}} {log}\left(\mathrm{3}\right)+\mathrm{4}^{\mathrm{2}} {log}\left(\mathrm{4}\right)+…+{n}^{\mathrm{2}} {log}\left({n}\right) \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}{log}\left({n}\right)−\frac{{n}^{\mathrm{3}} }{\mathrm{9}}+\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{4}\pi^{\mathrm{2}} }+\frac{{n}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{360}{n}}+…+{C}\:\: \\ $$$${But}\:{i}\:{haven}'{t}\:{found}\:{any}\:{proof}\: \\ $$

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