study the convergence of Σ (1/(nH_n )) with H_n =Σ


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Question Number 74795 by mathmax by abdo last updated on 30/Nov/19

$s t u d y t h e c o n v e r g e n c e o f Σ \frac{1}{{n H}_{n}}$ $w i t h H_{n} = \sum_{k = 1}^{n} \frac{1}{k}$
Commented by mathmax by abdo last updated on 01/Dec/19

$w e h a v e H_{n} \sim l n (n) (n \to + \infty) \Rightarrow \frac{1}{n H_{n}} \sim \frac{1}{n l n (n)}$ $l e t U_{n} = \frac{1}{n l n (n)} w e h a v e U_{n} d e c r e a s e t o 0 s o \int_{2}^{+ \infty} \frac{d t}{t l n (t)} a n d$ $Σ U_{n} h a v e s a m e n a t u r e c h a n g e m e n t l n (t) = u g i v e$ $\int_{2}^{+ \infty} \frac{d t}{t l n (t)} = \int_{l n (2)}^{+ \infty} \frac{e^{u} d u}{e^{u} u} = \int_{l n (2)}^{+ \infty} \frac{d u}{u} = + \infty \Rightarrow Σ U_{n} d i v e r g e s \Rightarrow$ $Σ \frac{1}{{n H}_{n}} d i v e r g e s .$
	Answered by mind is power last updated on 01/Dec/19

	$Σ \frac{1}{{nH}_{n}}$ $H_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{k}$ $\frac{1}{k + 1} ⩽ \int_{k}^{k + 1} \frac{1}{t} ⩽ \frac{1}{k}$ $\Rightarrow \underset{k = 1}{\sum^{n}} \frac{1}{k + 1} ⩽ \ln (n + 1) ⩽ H_{n}$ $\Rightarrow H_{n} ⩾ \ln (n + 1) \Rightarrow \frac{1}{{nH}_{n}} ⩽ \frac{1}{nln (n + 1)}$ $H_{n + 1} - 1 ⩽ \ln (n + 1)$ $H_{n} ⩽ \ln (n) + 1 \Rightarrow \frac{1}{{nH}_{n}} ⩾ \frac{1}{nln (n) + n}$ $\Rightarrow Σ \frac{1}{{nH}_{n}} ⩾ \sum_{n ⩾ 1} \frac{1}{n (\ln (n) + 1)} \to + \infty$
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