Question Number 4392 by Rasheed Soomro last updated on 18/Jan/16
$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}}\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{n}}} =?\:\:\:,\:\mathrm{x}>\mathrm{1} \\ $$
Commented by Yozzii last updated on 18/Jan/16
$$ \\ $$$$\frac{\mathrm{1}}{{n}}{x}^{\mathrm{1}/{n}} =\frac{\mathrm{1}}{{n}}{x}.{x}^{{n}^{−\mathrm{1}} −\mathrm{1}} ={x}\frac{{d}}{{dx}}\left({x}^{{n}^{−\mathrm{1}} } \right) \\ $$$$\therefore\:{S}\left({N}\right)={x}\underset{{n}=\mathrm{1}} {\overset{{N}} {\sum}}\frac{{d}}{{dx}}\left({x}^{\mathrm{1}/{n}} \right) \\ $$$${S}\left({N}\right)={x}\frac{{d}}{{dx}}\left(\underset{{n}=\mathrm{1}} {\overset{{N}} {\sum}}{x}^{\mathrm{1}/{n}} \right) \\ $$$$ \\ $$$${I}=\int_{\mathrm{1}} ^{\infty} {x}^{\mathrm{1}/{v}} {dv}. \\ $$$${let}\:{x}^{\mathrm{1}/{v}} ={e}^{−{u}} \therefore\:{v}^{−\mathrm{1}} {lnx}=−{u} \\ $$$$\therefore−{v}^{−\mathrm{2}} {lnxdv}=−{du} \\ $$$${dv}={v}^{\mathrm{2}} \left({lnx}\right)^{−\mathrm{1}} {du} \\ $$$${v}=−{u}^{−\mathrm{1}} {lnx}\therefore\:{v}^{\mathrm{2}} ={u}^{−\mathrm{2}} {ln}^{\mathrm{2}} {x} \\ $$$${dv}={u}^{−\mathrm{2}} {ln}^{\mathrm{2}} {x}×\left({lnx}\right)^{−\mathrm{1}} {du} \\ $$$${dv}={u}^{−\mathrm{2}} \left({lnx}\right){du} \\ $$$${At}\:{v}=\mathrm{1}\Rightarrow{u}=−{lnx} \\ $$$${As}\:{v}\rightarrow\infty\:{u}\rightarrow\mathrm{0} \\ $$$$\therefore{I}=\int_{−{lnx}} ^{\mathrm{0}} {e}^{−{u}} ×{u}^{−\mathrm{2}} {lnxdu} \\ $$$${I}=\left({lnx}\right)\int_{−{lnx}} ^{\mathrm{0}} {u}^{−\mathrm{2}} {e}^{−{u}} {du} \\ $$$$ \\ $$$$ \\ $$