Question Number 126977 by Dwaipayan Shikari last updated on 25/Dec/20
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{7}} }{\mathrm{7}^{{n}} } \\ $$
Commented by MJS_new last updated on 25/Dec/20
$$=\frac{\mathrm{285929}}{\mathrm{11664}} \\ $$
Answered by mindispower last updated on 25/Dec/20
$$\frac{\mathrm{1}}{\mathrm{7}^{{n}} }={e}^{−{nln}\left(\mathrm{7}\right)} \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{7}} }{\mathrm{7}^{{n}} } \\ $$$${let}\:{f}_{{n}} \left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{{nx}} ,{x}<\mathrm{0} \\ $$$${f}_{{n}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{e}^{{x}} } \\ $$$${g}_{{n}} \left({x}\right)=\frac{\partial^{\mathrm{7}} }{\partial{x}^{\mathrm{7}} }{f}_{{n}} \left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{n}^{\mathrm{7}} {e}^{{nx}} \\ $$$${g}_{{n}} \left(−{ln}\left(\mathrm{7}\right)\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}{n}^{\mathrm{7}} .\frac{\mathrm{1}}{\mathrm{7}^{{n}} }=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{n}^{\mathrm{7}} }{\mathrm{7}^{{n}} }={S} \\ $$