Use-the-limit-comparison-test-to-determine-if-the-series-converges-or-diverges-n-2-1-7-8n-ln-ln-n-

Question Number 140825 by liberty last updated on 13/May/21

Use the limit comparison test

to determine if the series converges

or diverges

\underset{n = 2}{\sum^{\infty}} \frac{1}{7 + 8 n \ln (\ln n)} .

Answered by mathmax by abdo last updated on 13/May/21

let φ (x) = \frac{1}{7 + 8 nln (lnn)} (n > 2) \Rightarrow φ^{^{'}} (x) = - \frac{{(7 + 8 xlog (logx))}^{^{'}}}{(7 + 8 xlog {(logx)}^{2}}

= - \frac{8 \log (logx) + 8 x \times \frac{1}{xlogx}}{{(\dots .)}^{2}} = - \frac{8 \log (logx) + \frac{8}{logx}}{{(\dots .)}^{2}} < 0 \Rightarrow φ is decreazing

what about \int_{e}^{\infty} \frac{dx}{7 + 8 x \log (logx)} ? changement logx = t give

\int_{e}^{\infty} \frac{dx}{7 + 8 xlog (logx)} = \int_{1}^{\infty} \frac{e^{t} dt}{7 + {8 e}^{t} \log (t)} = \int_{1}^{\infty} \frac{dt}{{7 e}^{- t} + 8 logt}

at + \infty \frac{1}{{7 e}^{- t} + 8 logt} \sim \frac{1}{8 logt} and

\int_{1}^{\infty} \frac{dt}{logt} =_{logt = u} \int_{0}^{\infty} \frac{e^{u}}{u} du \lim_{u \to + \infty} \sqrt{u} . \frac{e^{u}}{u} = + \infty \Rightarrow this integral diverges

\Rightarrow Σ \frac{1}{7 + 8 nlog (logn)} diverges \dots!

Commented by liberty last updated on 13/May/21

why sir \int_{e}^{\infty} \frac{dx}{7 + 8 x \ln (\ln x)} ?

it should be \int_{2}^{\infty} \dots dx sir ?

Commented by Mathspace last updated on 13/May/21

\int_{2}^{\infty} \dots d x a n d \int_{e}^{+ \infty} \dots d x a r e t h e s a m e

f o r c o n v e r g e n c e (n o t v a l u e!)