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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 \underset{n = 2}{\sum^{\infty}} \frac{1}{n^{p} . {(\ln n)}^{q}} ?

(a) always converges

(b) always diverges

(c) may converges or diverges

Answered by mnjuly1970 last updated on 27/Feb/21

c a u c h y d e n s i t y t e s t .

\underset{k = 1}{\sum^{\infty}} 2^{k} \frac{1}{2^{k p} {(l n 2)}^{q} k^{q}} = \frac{1}{{(l n 2)}^{q}} \underset{k = 1}{\sum^{\infty}} \frac{1}{2^{k (p - 1)} k^{q}}

\sum_{k = 1} {(\frac{1}{2^{p - 1}})}^{k} {(\frac{1}{k})}^{q} \dots f o r (p > 1, q > 1) i s c o n v e r g e n t

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

the sequence u_{n} = \frac{1}{n^{p} {(logn)}^{q}} decreaze to 0 the serie have nature [of

\int_{e}^{\infty} \frac{dt}{t^{p} {(logt)}^{q}} =_{logt = u} \int_{1}^{\infty} \frac{e^{u} du}{e^{pu} u^{q}} = \int_{1}^{\infty} e^{(1 - p) u} \frac{du}{u^{q}}

we have \lim_{u \to + \infty} u^{2} \frac{e^{- (p - 1) u}}{u^{q}} = \lim_{u \to + \infty} u^{2 - q} e^{- (p - 1) u} = 0 \Rightarrow

the integral is convergent \Rightarrow the serie is cv .