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Question Number 41410 by maxmathsup by imad last updated on 06/Aug/18

l e t A_{n} = \int_{0}^{\infty} [{n e}^{- x}] d x w i t h n ⩾ 2

1) c a l c u l a t e A_{n}

2) f i n d n a t u r e o f \sum_{n ⩾ 2} A_{n}

3) s t u d y t h e c o n v e r g e n c e o f Σ \frac{1}{A_{n}} a n d Σ \frac{1}{A_{n}^{2}}

Commented by maxmathsup by imad last updated on 07/Aug/18

1) c h a n g e m e n t n e^{- x} = t g i v e e^{- x} = \frac{t}{n} \Rightarrow e^{x} = \frac{n}{t} \Rightarrow x = l n (n) - l n (t) \Rightarrow

d x = - \frac{d t}{t} \Rightarrow A_{n} = - \int_{n}^{0} [t] \frac{d t}{t} = \int_{0}^{n} \frac{[t]}{t} d t = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} \frac{k}{t} d t

= \sum_{k = 0}^{n - 1} k {l n (k + 1) - l n (k)} = \sum_{k = 1}^{n - 1} k {l n (k + 1) - l n (k)} \Rightarrow

A_{n} = \sum_{k = 1}^{n - 1} k l n (1 + \frac{1}{k})

Commented by maxmathsup by imad last updated on 07/Aug/18

2) w e h a v e {l n}^{^{'}} (1 + x) = \frac{1}{1 + x} = 1 - x + x^{2} - \dots i f ∣ x ∣< 1 \Rightarrow

l n (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \dots \Rightarrow l n (1 + x) ⩾ x - \frac{x^{2}}{2} \Rightarrow l n (1 + \frac{1}{k}) ⩾ \frac{1}{k} - \frac{1}{2 k^{2}} \Rightarrow

\forall k \in [[0, n - 1]] k l n (1 + \frac{1}{k}) ⩾ 1 - \frac{1}{2 k} \Rightarrow \sum_{k = 1}^{n - 1} k l n (1 + \frac{1}{k}) ⩾ \sum_{k = 1}^{n - 1} (1 - \frac{1}{2 k}) \Rightarrow

A_{n} ⩾ n - 1 - \frac{1}{2} \sum_{k = 1}^{n - 1} \frac{1}{k} \Rightarrow A_{n} ⩾ n - 1 - \frac{1}{2} H_{n - 1} b u t

H_{n - 1} = l n (n - 1) + γ + o (\frac{1}{n}) \Rightarrow n - 1 - \frac{1}{2} H_{n - 1} = n - 1 - l n \sqrt{n - 1} - \frac{γ}{2} + o (\frac{1}{n})

b u t {l i m}_{n \to + \infty} n - 1 - l n \sqrt{n - 1} = + \infty \Rightarrow A_{n} \to + \infty \Rightarrow Σ A_{n} d i v e r g e s