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Question Number 216907 by MrGaster last updated on 24/Feb/25

$$\mathrm{Prove}:n!=1+\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{k^n}{e^k}-\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{B_k\sin \left(\pi k\right)(n-k)!}{\pi k}$$

Answered by MrGaster last updated on 13/Mar/25

$$\mathrm{Let}f\left(x\right)=\frac{x}{e^x-1}\stackrel{\iff }{\mathrm{\&}}\frac{x}{e^x-1}=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{B_k}{k!}{x}^k$$$$n!\mathrm{\&}n!=\mathrm{\Gamma }(n+1)\Rightarrow \mathrm{\Gamma }(n+1)={\int }_0^{\mathrm{\infty }}{t}^n{e}^{-t}dt$$$$\mathrm{Using}\mathrm{the}\mathrm{Euler}-\mathrm{Maclaurinr}$$$$\Rightarrow \underset{k=1}{\sum ^{\mathrm{\infty }}}f\left(k\right)={\int }_1^{\mathrm{\infty }}f\left(x\right)dx+\frac{f\left(1\right)}{2}+\underset{k=1}{\sum ^m}\frac{B_{2k}}{\left(2k\right)!}{f}^{(2k+1)}\left(1\right)+R_m$$$$f\left(x\right)=\frac{x^n}{e^x}\Rightarrow \underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{k^n}{e^k}={\int }_1^{\mathrm{\infty }}\frac{x^n}{e^x}dx+\frac{1}{2e}+R_m\stackrel{\iff }{\mathrm{\&}}f\left(x\right)=\frac{B_k\sin \left(\pi k\right)(n-k)!}{\pi k}\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{B_k\sin \left(\pi k\right)(n-k)!}{\pi k}={\int }_1^{\mathrm{\infty }}\frac{B_k\sin \left(\pi x\right)(n-x)!}{\pi x}dx+\frac{B_1\sin \left(\pi \right)(n-1)!}{\pi }+\underset{k=1}{\sum ^m}\frac{B_{2k}}{\left(2k\right)!}\frac{d^{2k-1}}{{dx}^{2k-1}}\left(\frac{B_k\sin \left(\pi x\right)(n-x)!}{\pi x}\right){\mid }_{x=1}+R_m$$$$\mathrm{Combining}\mathrm{these}\mathrm{results}\Rightarrow n!=1+\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{k^n}{e^k}-\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{B_k\sin \left(\pi k\right)(n-k)!}{\pi k}$$$$[\mathrm{Q}.\mathrm{E}.\mathrm{D}]$$

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