let S_n = 1 +(1/((^3 (√2)))) + (1/((^3 (√3)))) + ....+(1/((^3 (√n)))) calculate lim _(n→+∞) S


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Question Number 41517 by maxmathsup by imad last updated on 08/Aug/18

$l e t S_{n} = 1 + \frac{1}{(^{3} \sqrt{2})} + \frac{1}{(^{3} \sqrt{3})} + . . . . + \frac{1}{(^{3} \sqrt{n})}$ $c a l c u l a t e l i m_{n \to + \infty} S_{n}$
Commented by maxmathsup by imad last updated on 10/Aug/18
$we have S_n =Σ_(k=1) ^n (1/k^(1/3) ) the function t→(1/x^(1/3) ) is decreasing so ∫_k ^(k+1) (dt/t^(1/3) ) ≤ (1/k^(1/3) ) ≤ ∫_(k−1) ^k (dt/t^(1/3) ) ⇒ Σ_(k=2) ^n ∫_k ^(k+1) (dt/t^(1/3) ) ≤ Σ_(k=2) ^n (1/k^(1/3) ) ≤ Σ_(k=2) ^n ∫_(k−1) ^k (dt/t^(1/3) ) ⇒ ∫_2 ^(n+1) t^(−(1/3)) dt ≤ S_n −1 ≤ ∫_1 ^n t^(−(1/3)) dt ⇒ [ (3/2) t^(2/3) ]_2 ^(n+1) ≤ S_n −1 ≤ [ (3/2) t^(2/3) ]_1 ^n ⇒(3/2){ (n+1)^(2/3) −2^(2/3) }≤ S_n −1 ≤ (3/2){ n^(2/3) −1}⇒ S_n ≥ 1 +(3/2)(n+1)^(2/3) −(3/2) 2^(2/3) →+∞(n→+∞) ⇒ lim_(n→+∞) S_n =+∞ .$
$w e h a v e S_{n} = \sum_{k = 1}^{n} \frac{1}{k^{\frac{1}{3}}} t h e f u n c t i o n t \to \frac{1}{x^{\frac{1}{3}}} i s d e c r e a s i n g s o$ $\int_{k}^{k + 1} \frac{d t}{t^{\frac{1}{3}}} ⩽ \frac{1}{k^{\frac{1}{3}}} ⩽ \int_{k - 1}^{k} \frac{d t}{t^{\frac{1}{3}}} \Rightarrow \sum_{k = 2}^{n} \int_{k}^{k + 1} \frac{d t}{t^{\frac{1}{3}}} ⩽ \sum_{k = 2}^{n} \frac{1}{k^{\frac{1}{3}}} ⩽ \sum_{k = 2}^{n} \int_{k - 1}^{k} \frac{d t}{t^{\frac{1}{3}}}$ $\Rightarrow \int_{2}^{n + 1} t^{- \frac{1}{3}} d t ⩽ S_{n} - 1 ⩽ \int_{1}^{n} t^{- \frac{1}{3}} d t \Rightarrow$ ${[\frac{3}{2} t^{\frac{2}{3}}]}_{2}^{n + 1} ⩽ S_{n} - 1 ⩽ {[\frac{3}{2} t^{\frac{2}{3}}]}_{1}^{n} \Rightarrow \frac{3}{2} {{(n + 1)}^{\frac{2}{3}} - 2^{\frac{2}{3}}} ⩽ S_{n} - 1$ $⩽ \frac{3}{2} {n^{\frac{2}{3}} - 1} \Rightarrow S_{n} ⩾ 1 + \frac{3}{2} {(n + 1)}^{\frac{2}{3}} - \frac{3}{2} 2^{\frac{2}{3}} \to + \infty (n \to + \infty) \Rightarrow$ ${l i m}_{n \to + \infty} S_{n} = + \infty .$
	Answered by alex041103 last updated on 09/Aug/18

	$\underset{n \to \infty}{l i m} S_{n} = S_{\infty} = \underset{k = 1}{\sum^{\infty}} \frac{1}{k^{1 / 3}}$ $W e k n o w t h a t k^{1 / 3} ⩽ k f o r k ϵ N$ $\Rightarrow \frac{1}{k^{1 / 3}} ⩾ \frac{1}{k}$ $\Rightarrow S_{\infty} > \underset{k = 1}{\sum^{\infty}} \frac{1}{k} = s_{\infty}$ $A s w e k n o w s_{\infty} {d o e s n}^{'} t c o n v e r g e .$ $\Rightarrow S_{\infty} {d o e s n}^{'} t c o n v e r g e .$
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