n-1-log-n-n-2-Does-the-series-converge-or-diverge-help-find-the-sum-

Question Number 46527 by Tawa1 last updated on 28/Oct/18

\underset{n = 1}{\sum^{\infty}} {(\frac{\log n}{n})}^{2}

Does the series converge or diverge, help find the sum \dots

Commented by maxmathsup by imad last updated on 28/Oct/18

l e t φ (x) = {(\frac{l n (x)}{x})}^{2} w i t h x ⩾ 2 w e h a v e φ^{^{'}} (x) = 2 \frac{l n (x)}{x} {(\frac{l n (x)}{x})}^{^{'}}

= \frac{2 l n x}{x} {\frac{1 - l n (x)}{x^{2}}} < 0 φ i s d e c r e a s i n g o n [2, + \infty [s o \sum_{n = 1}^{\infty} {(\frac{l n (n)}{n})}^{2} a n d

\int_{2}^{+ \infty} {(\frac{l n (x)}{x})}^{2} d x h a v e t h e s a m e n a t u r e o f c o n v e r g e n c e b u t

\int_{2}^{+ \infty} \frac{{(l n (x))}^{2}}{x^{2}} d x =_{l n (x) = t} \int_{l n (2)}^{) \infty} \frac{t^{2}}{e^{2 t}} e^{t} d t = \int_{l n (2)}^{+ \infty} t^{2} e^{- t} d t a n d t h i s i n t e g r a l

c o n v e r g e s \Rightarrow \sum_{n = 1}^{\infty} {(\frac{l n (n)}{n})}^{2} c o n v e r g e s .

Commented by MJS last updated on 28/Oct/18

it seems that \underset{1}{\int^{+ \infty}} {(\frac{\ln x}{x})}^{2} d x = 2

Commented by Tawa1 last updated on 28/Oct/18

God bless you sir

Commented by Tawa1 last updated on 28/Oct/18

Please how can i prove it is monotonic first sir

Commented by math khazana by abdo last updated on 28/Oct/18

t h a n k y o u s i r .