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Question Number 146735 by qaz last updated on 15/Jul/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{\mathrm{n}}}{(\mathrm{n}+1)(\mathrm{n}+2)}=?$$

Answered by Olaf_Thorendsen last updated on 15/Jul/21

$${\mathrm{S}}_{\mathrm{N}}=\underset{n=1}{\sum ^{\mathrm{N}}}\frac{H_n}{n+1}-\underset{n=1}{\sum ^{\mathrm{N}}}\frac{H_n}{n+2}$$$${\mathrm{S}}_{\mathrm{N}}=\underset{n=1}{\sum ^{\mathrm{N}}}\frac{H_n+\frac{1}{n+1}}{n+1}-\underset{n=1}{\sum ^{\mathrm{N}}}\frac{1}{{(n+1)}^2}$$$$-\underset{n=1}{\sum ^{\mathrm{N}}}\frac{H_n+\frac{1}{n+1}+\frac{1}{n+2}}{n+2}+\underset{n=1}{\sum ^{\mathrm{N}}}\frac{1}{(n+1)(n+2)}$$$$+\underset{n=1}{\sum ^{\mathrm{N}}}\frac{1}{{(n+2)}^2}$$$$$$$${\mathrm{S}}_{\mathrm{N}}=\underset{n=2}{\sum ^{\mathrm{N}+1}}\frac{H_n}{n}-\underset{n=2}{\sum ^{\mathrm{N}+1}}\frac{1}{n^2}-\underset{n=3}{\sum ^{\mathrm{N}+2}}\frac{H_n}{n}$$$$+\underset{n=2}{\sum ^{\mathrm{N}+1}}\frac{1}{n}-\underset{n=3}{\sum ^{\mathrm{N}+2}}\frac{1}{n}+\underset{n=3}{\sum ^{\mathrm{N}+2}}\frac{1}{n^2}$$$${\mathrm{S}}_{\mathrm{N}}=\frac{H_2}{2}-\frac{H_{n+2}}{n+2}+\frac{1}{{(n+2)}^2}-\frac{1}{2^2}+\frac{1}{2}-\frac{1}{n+2}$$$$$$$$H_n\sim \ln n+\gamma \Rightarrow \frac{H_{n+2}}{n+2}\to 0$$$$$$$$\mathrm{Finally}{\mathrm{S}}_{\mathrm{\infty }}=\frac{1+\frac{1}{2}}{2}-\frac{1}{4}+\frac{1}{2}=1$$

Answered by mnjuly1970 last updated on 15/Jul/21

$$solution..$$$$weknowthat:$$$$\sum _{n⩾1}\frac{H_n{x}^{n+1}}{n+1}=\frac{1}{2}{log}^2(1-x)$$$$\sum _{n⩾1}\frac{H_n}{(n+1)(n+2)}=\frac{1}{2}{\int }_0^1{log}^2\left(x\right)dx$$$$=\frac{1}{2}\{{\left[{xlog}^2\left(x\right)\right]}_0^1-2{\int }_0^{1}log\left(x\right)dx\}$$$$=-{\int }_0^{1}log\left(x\right)dx=1....✓$$$$$$$$$$

Commented by qaz last updated on 15/Jul/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{H}}_{\mathrm{n}}}{(\mathrm{n}+1)(\mathrm{n}+2)}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{\mathrm{H}}_{\mathrm{n}}{\int }_0^1({\mathrm{x}}^{\mathrm{n}}-{\mathrm{x}}^{\mathrm{n}+1})\mathrm{dx}={\int }_0^1(-\frac{\ln (1-\mathrm{x})}{1-\mathrm{x}}+\frac{\mathrm{xln}(1-\mathrm{x})}{1-\mathrm{x}})\mathrm{dx}$$$$=-{\int }_0^1\ln (1-\mathrm{x})\mathrm{dx}=1$$$$---------------$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{\mathrm{H}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{k}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{n}=\mathrm{k}}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{k}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^{\mathrm{n}+\mathrm{k}}}{\mathrm{k}}=\left(\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^{\mathrm{k}}}{\mathrm{k}}\right)\left(\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{x}}^{\mathrm{n}}\right)=-\frac{\ln (1-\mathrm{x})}{1-\mathrm{x}}$$

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