let u_n =Σ_(k=1) ^n (((−1)^([k]) )/k) and H_n =Σ_(k=1) ^n (1/k) 1)calculate u_n interms of H_n 2)study the convergence of (u_n ) 3)study theconvergence of Σ u


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Question Number 45792 by maxmathsup by imad last updated on 16/Oct/18

$l e t u_{n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{[k]}}{k} a n d H_{n} = \sum_{k = 1}^{n} \frac{1}{k}$ $1) c a l c u l a t e u_{n} i n t e r m s o f H_{n}$ $2) s t u d y t h e c o n v e r g e n c e o f (u_{n})$ $3) s t u d y t h e c o n v e r g e n c e o f Σ u_{n .}$
Commented by maxmathsup by imad last updated on 17/Oct/18

$3) {l i m}_{n \to + \infty} u_{n} \neq 0 \Rightarrow Σ u_{n} d i v e r g e s . .$
Commented by maxmathsup by imad last updated on 17/Oct/18

$2) c o n v e r g e n c e o f (u_{n}) ?$ $w e h a v e u_{2 n} = \frac{1}{2} H_{n} + \frac{1}{2} H_{n - 1} - H_{2 n - 1}$ $= \frac{1}{2} (l n (n) + γ + o (\frac{1}{n})) + \frac{1}{2} (l n (n - 1) + γ + o (\frac{1}{n})) - l n (2 n - 1) - γ - o (\frac{1}{n})$ $= l n \sqrt{n (n + 1)} - l n (2 n - 1) + o (\frac{1}{n}) = l n (\frac{\sqrt{n^{2} + n}}{2 n - 1}) + o (\frac{1}{n}) \to - l n (2) (n \to + \infty)$ $u_{2 n + 1} = \frac{1}{2} H_{n} + \frac{1}{2} H_{n} - H_{2 n + 1} = H_{n} - H_{2 n + 1}$ $= l n (n) + γ - l n (2 n + 1) - γ + o (\frac{1}{n}) = l n (\frac{n}{2 n + 1}) + o (\frac{1}{n}) \to - l n (2) (n \to + \infty)$ $s o (u_{n}) c o n v e r g e s a n d {l i m}_{n \to + \infty} u_{n} = - l n (2) .$
Commented by maxmathsup by imad last updated on 17/Oct/18

$w e h a v e u_{n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{[k]}}{k} \Rightarrow u_{n} = \sum_{p = 1}^{[\frac{n}{2}]} \frac{1}{2 p} - \sum_{p = 0}^{[\frac{n - 1}{2}]} \frac{1}{2 p + 1} b u t$ $\sum_{p = 1}^{[\frac{n}{2}]} \frac{1}{2 p} = \frac{1}{2} H_{[\frac{n}{2}]}$ $\sum_{p = 0}^{[\frac{n - 1}{2}]} \frac{1}{2 p + 1} = 1 + \frac{1}{3} + \frac{1}{5} + . . . . + \frac{1}{2 [\frac{n - 1}{2}] + 1}$ $= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . . . + \frac{1}{2 [\frac{n - 1}{2}]} + \frac{1}{2 [\frac{n - 1}{2}] + 1} - \frac{1}{2} (1 + \frac{1}{2} + . . . . + \frac{1}{2 [\frac{n - 1}{2}]})$ $= H_{2 [\frac{n - 1}{2}] + 1} - \frac{1}{2} H_{[\frac{n - 1}{2}]} \Rightarrow$ $u_{n} = \frac{1}{2} H_{[\frac{n}{2}]} + \frac{1}{2} H_{[\frac{n - 1}{2}]} - H_{2 [\frac{n - 1}{2}] + 1}$
	Answered by arcana last updated on 17/Oct/18

	$1) c o m o u_{n} = \underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k}}{k}$ $s i n = 2 t e s p a r (p a r a a l g u n t e n t e r o p o s i t i v o)$ $u_{n} = \underset{k = 1}{\sum^{n / 2}} \frac{{(- 1)}^{2 k - 1}}{2 k - 1} + \underset{k = 1}{\sum^{n / 2}} \frac{{(- 1)}^{2 k}}{2 k}$ $= - \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1} + \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k}$ $= - \underset{k = 1}{\sum^{n / 2}} \frac{1}{2 k - 1} + \frac{1}{2} H_{n / 2}$ $t o m a n d o m = 2 k - 1, l u e g o$ $= - \underset{m = 1}{\sum^{n - 1}} \frac{1}{m} + \frac{1}{2} H_{n / 2} = \frac{1}{2} H_{n / 2} - H_{n - 1}$ $s i n = 2 t - 1 e s i m p a r, t e n e m o s$ $u_{n} = \underset{k = 1}{\sum^{(n + 1) / 2}} \frac{{(- 1)}^{2 k - 1}}{2 k - 1} + \underset{k = 1}{\sum^{(n - 1) / 2}} \frac{{(- 1)}^{2 k}}{2 k}$ $= - \underset{k = 1}{\sum^{(n + 1) / 2}} \frac{1}{2 k - 1} + \underset{k = 1}{\sum^{(n - 1) / 2}} \frac{1}{2 k}$ $= - \underset{k = 1}{\sum^{(n + 1) / 2}} \frac{1}{2 k - 1} + \frac{1}{2} H_{(n - 1) / 2}$ $t o m a n d o m = 2 k - 1$ $= - \underset{m = 1}{\sum^{n}} \frac{1}{m} + \frac{1}{2} H_{(n - 1) / 2} = - H_{n} + \frac{1}{2} H_{(n - 1) / 2}$
	Commented by maxmathsup by imad last updated on 17/Oct/18

	$t h a n k y o u A r k a n a f o r t h i s w o r k . . .$
	Answered by arcana last updated on 17/Oct/18

	$3 . c o m o \frac{1}{1 + x} = \underset{k = 0}{\sum^{\infty}} {(- x)}^{k} p a r a ∣ x ∣< 1$ $i n t e g r a n d o a a m b o s l a d o s v e m o s q u e$ $\Rightarrow l n (1 + x) = \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k + 1}}{k + 1} \cdot x^{k + 1}$ $c o n - 1 < x ⩽ 1 . S i x = 1, e n t o n c e s$ $l n (2) = \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k + 1}}{k + 1} = Σ u_{n}$
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