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Question Number 78799 by Pratah last updated on 20/Jan/20

Commented by mathmax by abdo last updated on 20/Jan/20

$$convergence?letf\left(x\right)=\frac{ln\left(x\right)}{x^{\frac{5}{4}}}withx⩾2wehave$$$$f^{{}^{\prime }}\left(x\right)=\frac{\frac{1}{x}{x}^{\frac{5}{4}}-\frac{5}{4}{x}^{\frac{5}{4}-1}lnx}{x^{\frac{5}{2}}}=\frac{x^{\frac{5}{4}-1}-\frac{5}{4}{x}^{\frac{5}{4}-1}lnx}{x^{\frac{5}{2}}}$$$$=\frac{x^{\frac{1}{4}}}{x^{\frac{5}{2}}}\times (1-\frac{5}{4}lnx)=\frac{1-\frac{5}{4}ln\left(x\right)}{x^{\frac{5}{2}-\frac{1}{4}}}=\frac{4-5lnx}{4x^{\frac{9}{4}}}$$$$4-5lnx<0\iff 4<5lnx\Rightarrow lnx>\frac{4}{5}\Rightarrow x>e^{\frac{4}{5}}$$$$soforx>e^{\frac{4}{5}}{f}^{{}^{\prime }}<0\Rightarrow fisdecreazing\Rightarrow \sum _{n=2}^{\mathrm{\infty }}\frac{lnn}{n^{\frac{5}{4}}}and$$$${\int }_2^{+\mathrm{\infty }}\frac{lnx}{x^{\frac{5}{4}}}dxhavethesamenatureofconv.changement$$$$ln\left(x\right)=tgive{\int }_2^{+\mathrm{\infty }}\frac{lnx}{x^{\frac{5}{4}}}dx={\int }_{ln\left(2\right)}^{+\mathrm{\infty }}\frac{t}{{\left(e^t\right)}^{\frac{5}{4}}}{e}^{t}dt$$$$={\int }_2^{+\mathrm{\infty }}t{e}^{t-\frac{5}{4}t}dt={\int }_2^{+\mathrm{\infty }}t{e}^{-\frac{t}{4}}dtandthisintegralisconvergengent$$$$\Rightarrow thisserieisconvergent.$$

Commented by mind is power last updated on 20/Jan/20

$$\mathrm{Nice}\mathrm{Sir}$$

Commented by msup trace by abdo last updated on 21/Jan/20

$$thankssir.$$

Answered by mind is power last updated on 20/Jan/20

$$\mathrm{Serie}\mathrm{Cv}\mathrm{you}\mathrm{Want}\mathrm{too}\mathrm{find}\mathrm{Sum}?$$$$$$

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