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n-1-1-n-x-1-n-x-gt-1-




Question Number 4392 by Rasheed Soomro last updated on 18/Jan/16
Σ_(n=1) ^∞ (1/n)x^(1/n) =?   , x>1
$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}}\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{n}}} =?\:\:\:,\:\mathrm{x}>\mathrm{1} \\ $$
Commented by Yozzii last updated on 18/Jan/16
  (1/n)x^(1/n) =(1/n)x.x^(n^(−1) −1) =x(d/dx)(x^n^(−1)  )  ∴ S(N)=xΣ_(n=1) ^N (d/dx)(x^(1/n) )  S(N)=x(d/dx)(Σ_(n=1) ^N x^(1/n) )    I=∫_1 ^∞ x^(1/v) dv.  let x^(1/v) =e^(−u) ∴ v^(−1) lnx=−u  ∴−v^(−2) lnxdv=−du  dv=v^2 (lnx)^(−1) du  v=−u^(−1) lnx∴ v^2 =u^(−2) ln^2 x  dv=u^(−2) ln^2 x×(lnx)^(−1) du  dv=u^(−2) (lnx)du  At v=1⇒u=−lnx  As v→∞ u→0  ∴I=∫_(−lnx) ^0 e^(−u) ×u^(−2) lnxdu  I=(lnx)∫_(−lnx) ^0 u^(−2) e^(−u) du
$$ \\ $$$$\frac{\mathrm{1}}{{n}}{x}^{\mathrm{1}/{n}} =\frac{\mathrm{1}}{{n}}{x}.{x}^{{n}^{−\mathrm{1}} −\mathrm{1}} ={x}\frac{{d}}{{dx}}\left({x}^{{n}^{−\mathrm{1}} } \right) \\ $$$$\therefore\:{S}\left({N}\right)={x}\underset{{n}=\mathrm{1}} {\overset{{N}} {\sum}}\frac{{d}}{{dx}}\left({x}^{\mathrm{1}/{n}} \right) \\ $$$${S}\left({N}\right)={x}\frac{{d}}{{dx}}\left(\underset{{n}=\mathrm{1}} {\overset{{N}} {\sum}}{x}^{\mathrm{1}/{n}} \right) \\ $$$$ \\ $$$${I}=\int_{\mathrm{1}} ^{\infty} {x}^{\mathrm{1}/{v}} {dv}. \\ $$$${let}\:{x}^{\mathrm{1}/{v}} ={e}^{−{u}} \therefore\:{v}^{−\mathrm{1}} {lnx}=−{u} \\ $$$$\therefore−{v}^{−\mathrm{2}} {lnxdv}=−{du} \\ $$$${dv}={v}^{\mathrm{2}} \left({lnx}\right)^{−\mathrm{1}} {du} \\ $$$${v}=−{u}^{−\mathrm{1}} {lnx}\therefore\:{v}^{\mathrm{2}} ={u}^{−\mathrm{2}} {ln}^{\mathrm{2}} {x} \\ $$$${dv}={u}^{−\mathrm{2}} {ln}^{\mathrm{2}} {x}×\left({lnx}\right)^{−\mathrm{1}} {du} \\ $$$${dv}={u}^{−\mathrm{2}} \left({lnx}\right){du} \\ $$$${At}\:{v}=\mathrm{1}\Rightarrow{u}=−{lnx} \\ $$$${As}\:{v}\rightarrow\infty\:{u}\rightarrow\mathrm{0} \\ $$$$\therefore{I}=\int_{−{lnx}} ^{\mathrm{0}} {e}^{−{u}} ×{u}^{−\mathrm{2}} {lnxdu} \\ $$$${I}=\left({lnx}\right)\int_{−{lnx}} ^{\mathrm{0}} {u}^{−\mathrm{2}} {e}^{−{u}} {du} \\ $$$$ \\ $$$$ \\ $$

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