If $\frac{1}{\mathrm{n}+1}+\frac{1}{2{(\mathrm{n}+1)}^2}+\frac{1}{3{(\mathrm{n}+1)}^3}+\dots \dots =\mathrm{\lambda }\left(\frac{1}{\mathrm{n}}-\frac{1}{2{\mathrm{n}}^2}+\frac{1}{3{\mathrm{n}}^3}-\dots \dots \right) \mathrm{then} \mathrm{\lambda }=$

$\begin{array}{rcl} & & \frac{1}{\mathrm{n}+1}+\frac{1}{2}\frac{1}{{(\mathrm{n}+1)}^2}+\frac{1}{3}\frac{1}{{(\mathrm{n}+1)}^3}+\dots \dots \\ & =& -\log \left(1-\frac{1}{\mathrm{n}+1}\right)\\ & =& -\log \left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)\\ & =& \log \left(\frac{\mathrm{n}+1}{\mathrm{n}}\right)\\ & =& \log \left(1+\frac{1}{\mathrm{n}}\right)\\ & =& \frac{1}{\mathrm{n}}-\frac{{\left({\displaystyle \frac{1}{\mathrm{n}}}\right)}^2}{2}+\frac{{\left({\displaystyle \frac{1}{\mathrm{n}}}\right)}^3}{3}-.......\\ \mathrm{\lambda }& =& 1\end{array}$