Question Number 135884 by Dwaipayan Shikari last updated on 16/Mar/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} ^{\left(\mathrm{7}\right)} }{{n}^{\mathrm{2}} }−\frac{{H}_{{n}} ^{\left(\mathrm{7}\right)} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}^{{m}} }={H}_{{n}} ^{\left({m}\right)} \\ $$
Answered by mindispower last updated on 17/Mar/21
$${H}_{{n}+\mathrm{1}} ^{\left(\mathrm{7}\right)} ={H}_{{n}} ^{\left(\mathrm{7}\right)} +\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{7}} } \\ $$$${S}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{H}_{{n}} ^{\left(\mathrm{7}\right)} }{{n}^{\mathrm{2}} }−\Sigma\frac{{H}_{{n}+\mathrm{1}} ^{\left(\mathrm{7}\right)} −\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{7}} }}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\Sigma\frac{{H}_{{n}} ^{\left(\mathrm{7}\right)} }{{n}^{\mathrm{2}} }−\frac{{H}_{{n}+\mathrm{1}} ^{\left(\mathrm{7}\right)} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{9}} } \\ $$$$=\frac{{H}_{\mathrm{1}} ^{\left(\mathrm{7}\right)} }{\mathrm{1}}+\zeta\left(\mathrm{9}\right)−\mathrm{1}=\zeta\left(\mathrm{9}\right) \\ $$