study the convergence of u_n =Σ_(k=2) ^n (1/(kln(k))) −ln(ln(n)


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Question Number 41706 by abdo.msup.com last updated on 11/Aug/18

$s t u d y t h e c o n v e r g e n c e o f$ $u_{n} = \sum_{k = 2}^{n} \frac{1}{k l n (k)} - l n (l n (n)$
Commented by math khazana by abdo last updated on 13/Aug/18
$the function f(x) =(1/(xln(x))) is decreasing on [2,+∞[ so ∫_k ^(k+1) f(x)dx≤ f(k)≤ ∫_(k−1) ^k f(x)dx⇒ ∫_k ^(k+1) (dx/(xln(x))) ≤ f(k)≤ ∫_(k−1) ^k f(x)dx but ∫_k ^(k+1) (dx/(xln(x))) =[ln(lnx)]_k ^(k+1) =ln(ln(k+1))−ln(ln(k)) ⇒ Σ_(k=3) ^n ∫_k ^(k+1) (dx/(xln(x)))=Σ_(k=3) ^n {ln(ln(k+1))−ln(ln(k))} =ln(ln(n+1))−ln(ln(3)) also Σ_(k=3) ^n ∫_(k−1) ^k (dx/(xln(x))) =∫_2 ^n (dx/(xln(x))) =[ln(ln(x))]_2 ^n ln(ln(n))−ln(ln(2)) ⇒ ln(ln(n+1))−ln(ln(3))≤ Σ_(k=3) ^n f(k) ≤ ln(ln(n))−ln(ln(2))⇒ ln(ln(n+1))−ln(ln(3)) +(1/(2ln(2))) −lnln(n)≤ u_n ≤−ln(ln(2)) +(1/(2ln(2))) ⇒ ln(((ln(n+1))/(ln(n))))−ln(ln(3))+(1/(2ln(2))) ≤ u_n ≤(1/(2ln(2)))−ln(ln(2)) so u_n is convergente and (1/(2ln(2))) −ln(ln(3))≤ lim_(n→+∞) u_n ≤ (1/(2ln(2))) −ln(ln2)$
$t h e f u n c t i o n f (x) = \frac{1}{x l n (x)} i s d e c r e a s i n g o n$ $[2, + \infty [s o \int_{k}^{k + 1} f (x) d x ⩽ f (k) ⩽ \int_{k - 1}^{k} f (x) d x \Rightarrow$ $\int_{k}^{k + 1} \frac{d x}{x l n (x)} ⩽ f (k) ⩽ \int_{k - 1}^{k} f (x) d x b u t$ $\int_{k}^{k + 1} \frac{d x}{x l n (x)} = {[l n (l n x)]}_{k}^{k + 1} = l n (l n (k + 1)) - l n (l n (k))$ $\Rightarrow \sum_{k = 3}^{n} \int_{k}^{k + 1} \frac{d x}{x l n (x)} = \sum_{k = 3}^{n} {l n (l n (k + 1)) - l n (l n (k))}$ $= l n (l n (n + 1)) - l n (l n (3)) a l s o$ $\sum_{k = 3}^{n} \int_{k - 1}^{k} \frac{d x}{x l n (x)} = \int_{2}^{n} \frac{d x}{x l n (x)} = {[l n (l n (x))]}_{2}^{n}$ $l n (l n (n)) - l n (l n (2)) \Rightarrow$ $l n (l n (n + 1)) - l n (l n (3)) ⩽ \sum_{k = 3}^{n} f (k)$ $⩽ l n (l n (n)) - l n (l n (2)) \Rightarrow$ $l n (l n (n + 1)) - l n (l n (3)) + \frac{1}{2 l n (2)} - l n l n (n) ⩽$ $u_{n} ⩽ - l n (l n (2)) + \frac{1}{2 l n (2)} \Rightarrow$ $l n (\frac{l n (n + 1)}{l n (n)}) - l n (l n (3)) + \frac{1}{2 l n (2)} ⩽ u_{n} ⩽ \frac{1}{2 l n (2)} - l n (l n (2))$ $s o u_{n} i s c o n v e r g e n t e a n d$ $\frac{1}{2 l n (2)} - l n (l n (3)) ⩽ {l i m}_{n \to + \infty} u_{n} ⩽ \frac{1}{2 l n (2)} - l n (l n 2)$
	Answered by alex041103 last updated on 12/Aug/18

	$W e c l e a r l y s e e t h a t \frac{d}{d k} (\frac{1}{k l n (k)}) < 0$ $W e a p p l y t h e i n t e g r a l t e s t :$ $u_{\infty} \approx \underset{n \to \infty}{l i m} [\int_{2}^{n} \frac{d k}{k l n (k)} - l n (l n (n))] =$ $= \underset{n \to \infty}{l i m} [\int_{2}^{n} \frac{d (l n (k))}{l n (k)} - l n (l n (n))] =$ $= \underset{n \to \infty}{l i m} [l n (l n (n)) - l n (l n (2)) - l n (l n (n))]$ $= - l n (l n (2))$ $A s w e k n o w l n \equiv {l o g}_{e} a n d e \approx 2.718$ $\Rightarrow S i n c e 1 < 2 < e 0 < l n (2) < 1$ $S i n c e l n (2) > 0 t h e n l n (l n (2)) e x i s t s .$ $\Rightarrow u_{n} c o n v e r g e s$
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