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If-p-gt-1-and-q-gt-1-what-can-be-said-about-the-convergence-of-n-2-1-n-p-ln-n-q-a-always-converges-b-always-diverges-c-may-converges-or-diverges-

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Question Number 134060 by benjo_mathlover last updated on 27/Feb/21
If p>1 and q>1 what can be said about the convergence of Σ_(n=2) ^∞ (1/(n^p .(ln n)^q )) ? (a) always converges (b) always diverges (c) may converges or diverges
$$\mathrm{If}\:\mathrm{p}>\mathrm{1}\:\mathrm{and}\:\mathrm{q}>\mathrm{1}\:\mathrm{what}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{said}\:\mathrm{about}\:\mathrm{the}\:\mathrm{convergence}\: \\ $$$$\mathrm{of}\:\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{p}} .\left(\mathrm{ln}\:\mathrm{n}\right)^{\mathrm{q}} }\:? \\ $$$$\left(\mathrm{a}\right)\:\mathrm{always}\:\mathrm{converges} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{always}\:\mathrm{diverges} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{may}\:\mathrm{converges}\:\mathrm{or}\:\mathrm{diverges} \\ $$
Answered by mnjuly1970 last updated on 27/Feb/21
cauchy density test. Σ_(k=1) ^∞ 2^k (1/(2^(kp) (ln2)^q k^q ))=(1/((ln2)^q ))Σ_(k=1) ^∞ (1/(2^(k(p−1)) k^q )) Σ_(k=1) ((1/2^(p−1) ))^k ((1/k))^q ...for (p>1, q>1) is convergent
$${cauchy}\:{density}\:{test}. \\ $$$$\:\:\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{kp}} \left({ln}\mathrm{2}\right)^{{q}} {k}^{{q}} }=\frac{\mathrm{1}}{\left({ln}\mathrm{2}\right)^{{q}} }\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{k}\left({p}−\mathrm{1}\right)} {k}^{{q}} } \\ $$$$\underset{{k}=\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{2}^{{p}−\mathrm{1}} }\right)^{{k}} \left(\frac{\mathrm{1}}{{k}}\right)^{{q}} \:\:…{for}\:\left({p}>\mathrm{1},\:\:{q}>\mathrm{1}\right)\:{is}\:{convergent} \\ $$
Answered by mathmax by abdo last updated on 27/Feb/21
the sequence u_n =(1/(n^p (logn)^q )) decreaze to0 the serie have nature[of ∫_e ^∞ (dt/(t^p (logt)^q )) =_(logt =u) ∫_1 ^∞ ((e^u du)/(e^(pu) u^q )) =∫_1 ^∞ e^((1−p)u) (du/u^q ) we have lim_(u→+∞) u^2 (e^(−(p−1)u) /u^q ) =lim_(u→+∞) u^(2−q) e^(−(p−1)u) =0 ⇒ the integral is convergent ⇒the serie is cv.
$$\mathrm{the}\:\mathrm{sequence}\:\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{p}} \left(\mathrm{logn}\right)^{\mathrm{q}} }\:\mathrm{decreaze}\:\mathrm{to0}\:\:\mathrm{the}\:\mathrm{serie}\:\mathrm{have}\:\mathrm{nature}\left[\mathrm{of}\right. \\ $$$$\int_{\mathrm{e}} ^{\infty} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{p}} \left(\mathrm{logt}\right)^{\mathrm{q}} }\:\:=_{\mathrm{logt}\:=\mathrm{u}} \:\:\:\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{e}^{\mathrm{u}} \:\mathrm{du}}{\mathrm{e}^{\mathrm{pu}} \mathrm{u}^{\mathrm{q}} }\:=\int_{\mathrm{1}} ^{\infty} \:\mathrm{e}^{\left(\mathrm{1}−\mathrm{p}\right)\mathrm{u}} \:\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{q}} } \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{lim}_{\mathrm{u}\rightarrow+\infty} \mathrm{u}^{\mathrm{2}} \:\frac{\mathrm{e}^{−\left(\mathrm{p}−\mathrm{1}\right)\mathrm{u}} }{\mathrm{u}^{\mathrm{q}} }\:=\mathrm{lim}_{\mathrm{u}\rightarrow+\infty} \:\:\mathrm{u}^{\mathrm{2}−\mathrm{q}} \:\mathrm{e}^{−\left(\mathrm{p}−\mathrm{1}\right)\mathrm{u}} \:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{convergent}\:\Rightarrow\mathrm{the}\:\mathrm{serie}\:\mathrm{is}\:\mathrm{cv}. \\ $$

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