Σ_(n=3) ^∝ 1/n(ln n)^2 is the function converg or diverg ? pleas help me


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Question Number 68188 by mhmd last updated on 06/Sep/19

$\sum_{n = 3}^{\propto} 1 / n {(l n n)}^{2} i s t h e f u n c t i o n c o n v e r g o r d i v e r g ? p l e a s h e l p m e$
Commented by Abdo msup. last updated on 07/Sep/19

$l e t φ (x) = \frac{1}{x {(l n x)}^{2}} w i t h x ⩾ 3 w e h a v e$ $φ^{^{'}} (x) = - \frac{{(l n x)}^{2} + x \times \frac{2 l n x}{x}}{x^{2} {(l n x)}^{4}} = - \frac{{(l n x)}^{2} + + 2 l n x}{x^{2} {(l n x)}^{4}} \Rightarrow$ $φ^{^{'}} (x) < 0 \Rightarrow φ i s i n c r e a s i n g o n [3, + \infty [$ $s o \sum_{n = 3}^{\infty} \frac{1}{n {(l n n)}^{2}} a n d \int_{3}^{+ \infty} \frac{d x}{x {(l n x)}^{2}} h a v e t h e s a m e$ $n s t u r e o f c o n v e r v e n c e c h a n g e m e n t l n x = t g i v e$ $\int_{3}^{+ \infty} \frac{d x}{x {(l n x)}^{2}} = \int_{l n (3)}^{+ \infty} \frac{e^{t} d t}{e^{t} t^{2}} d t = \int_{l n (3)}^{+ \infty} \frac{d t}{t^{2}}$ $= {[- \frac{1}{t}]}_{l n (3)}^{+ \infty} = \frac{1}{l n (3)} < + \infty \Rightarrow t h i s i n t e g r a l c o n v e r g e s$ $\Rightarrow t h e s e r i e c o n v e r g e s .$
Commented by Abdo msup. last updated on 07/Sep/19

$s o r r y φ i s d e c r e a z i n g o n [3, + \infty [$
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