let U_n =(1/(nH_n )) with H_n =Σ_(k=1) ^n (1/k) study the convergence of Σ_(n≥1) U_n 2) study the convergence of Σ_(n≥1) U


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Question Number 53778 by maxmathsup by imad last updated on 25/Jan/19

$l e t U_{n} = \frac{1}{{n H}_{n}} w i t h H_{n} = \sum_{k = 1}^{n} \frac{1}{k}$ $s t u d y t h e c o n v e r g e n c e o f \sum_{n ⩾ 1} U_{n}$ $2) s t u d y t h e c o n v e r g e n c e o f \sum_{n ⩾ 1} U_{n}^{2}$
Commented by maxmathsup by imad last updated on 18/Feb/19
$we have H_n =ln(n)+γ +o((1/n)) ⇒nH_n =nln(n) +nγ +o(1) ⇒ (1/(nH_n )) ∼ (1/(nln(n)+nγ +o(1))) ∼ (1/(n(ln(n)+γ))) the sequence n→(1/(n(ln(n)+γ))) is decreasing positive so Σ U_n and ∫_e ^(+∞) (dx/(x(ln(x)+γ))) have the same nature changement ln(x)=t give ∫_e ^(+∞) (dx/(x{ln(x)+γ))) =∫_1 ^(+∞) ((e^t dt)/(e^t (t+γ)))dt =∫_1 ^(+∞) (dt/(t+γ)) and this integral diverges ⇒Σ U_n is divergent ..$
$w e h a v e H_{n} = l n (n) + γ + o (\frac{1}{n}) \Rightarrow {n H}_{n} = n l n (n) + n γ + o (1) \Rightarrow$ $\frac{1}{{n H}_{n}} \sim \frac{1}{n l n (n) + n γ + o (1)} \sim \frac{1}{n (l n (n) + γ)} t h e s e q u e n c e n \to \frac{1}{n (l n (n) + γ)}$ $i s d e c r e a s i n g p o s i t i v e s o Σ U_{n} a n d \int_{e}^{+ \infty} \frac{d x}{x (l n (x) + γ)} h a v e t h e s a m e n a t u r e$ $c h a n g e m e n t l n (x) = t g i v e \int_{e}^{+ \infty} \frac{d x}{x {l n (x) + γ)} = \int_{1}^{+ \infty} \frac{e^{t} d t}{e^{t} (t + γ)} d t$ $= \int_{1}^{+ \infty} \frac{d t}{t + γ} a n d t h i s i n t e g r a l d i v e r g e s \Rightarrow Σ U_{n} i s d i v e r g e n t . .$
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