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Question Number 134060 by benjo_mathlover last updated on 27/Feb/21

$$\mathrm{If}\mathrm{p}>1\mathrm{and}\mathrm{q}>1\mathrm{what}\mathrm{can}\mathrm{be}$$ $$\mathrm{said}\mathrm{about}\mathrm{the}\mathrm{convergence}$$ $$\mathrm{of}\underset{\mathrm{n}=2}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{n}}^{\mathrm{p}}.{\left(\ln \mathrm{n}\right)}^{\mathrm{q}}}?$$ $$\left(\mathrm{a}\right)\mathrm{always}\mathrm{converges}$$ $$\left(\mathrm{b}\right)\mathrm{always}\mathrm{diverges}$$ $$\left(\mathrm{c}\right)\mathrm{may}\mathrm{converges}\mathrm{or}\mathrm{diverges}$$

Answered by mnjuly1970 last updated on 27/Feb/21

$$cauchydensitytest.$$ $$\underset{k=1}{\sum ^{\mathrm{\infty }}}{2}^k\frac{1}{2^{kp}{\left(ln2\right)}^q{k}^q}=\frac{1}{{\left(ln2\right)}^q}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^{k(p-1)}{k}^q}$$ $$\sum _{k=1}{\left(\frac{1}{2^{p-1}}\right)}^k{\left(\frac{1}{k}\right)}^q...for(p>1,q>1)isconvergent$$

Answered by mathmax by abdo last updated on 27/Feb/21

$$\mathrm{the}\mathrm{sequence}{\mathrm{u}}_{\mathrm{n}}=\frac{1}{{\mathrm{n}}^{\mathrm{p}}{\left(\mathrm{logn}\right)}^{\mathrm{q}}}\mathrm{decreaze}\mathrm{to}0\mathrm{the}\mathrm{serie}\mathrm{have}\mathrm{nature}[\mathrm{of}$$ $${\int }_{\mathrm{e}}^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^{\mathrm{p}}{\left(\mathrm{logt}\right)}^{\mathrm{q}}}{=}_{\mathrm{logt}=\mathrm{u}}{\int }_1^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{u}}\mathrm{du}}{{\mathrm{e}}^{\mathrm{pu}}{\mathrm{u}}^{\mathrm{q}}}={\int }_1^{\mathrm{\infty }}{\mathrm{e}}^{(1-\mathrm{p})\mathrm{u}}\frac{\mathrm{du}}{{\mathrm{u}}^{\mathrm{q}}}$$ $$\mathrm{we}\mathrm{have}{\lim }_{\mathrm{u}\to +\mathrm{\infty }}{\mathrm{u}}^2\frac{{\mathrm{e}}^{-(\mathrm{p}-1)\mathrm{u}}}{{\mathrm{u}}^{\mathrm{q}}}={\lim }_{\mathrm{u}\to +\mathrm{\infty }}{\mathrm{u}}^{2-\mathrm{q}}{\mathrm{e}}^{-(\mathrm{p}-1)\mathrm{u}}=0\Rightarrow$$ $$\mathrm{the}\mathrm{integral}\mathrm{is}\mathrm{convergent}\Rightarrow \mathrm{the}\mathrm{serie}\mathrm{is}\mathrm{cv}.$$

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