let-A-n-k-1-n-1-k-n-ln-1-k-n-calculate-lim-n-A-n-

Question Number 35688 by prof Abdo imad last updated on 22/May/18

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

c a l c u l a t e {l i m}_{n \to + \infty} A_{n}

Commented by prof Abdo imad last updated on 22/May/18

w e h a v e A_{n} = \sum_{k = 1}^{n} \frac{1}{n (\frac{k}{n} + 1)} l n (1 + \frac{k}{n})

= \frac{1}{n} \sum_{k = 1}^{n} \frac{l n (1 + \frac{k}{n})}{1 + \frac{k}{n}} s o A_{n} i s a R i e m a n s u m

{l i m}_{n \to + \infty} A_{n} = \int_{0}^{1} \frac{l n (1 + x)}{1 + x} d x = I

b y p a r t s I = {[{l n}^{2} (1 + x)]}_{0}^{1} - \int_{0}^{1} \frac{l n (1 + x)}{1 + x} d x

= {l n (2)}^{2} - I \Rightarrow 2 I = {l n (2)}^{2} \Rightarrow

I = \frac{{l n (2)}^{2}}{2} = {l i m}_{n \to + \infty} A_{n}

\Rightarrow

Answered by tanmay.chaudhury50@gmail.com last updated on 22/May/18

= \frac{l i m}{n \to \infty} \times \frac{1}{n} \times \frac{1}{1 + (\frac{k}{n})} \times l n (1 + \frac{k}{n})

= \int_{0}^{1} \frac{l n (1 + x)}{1 + x} \times d x

t = l n (1 + x) d t = \frac{d x}{1 + x}

\int_{0}^{l n 2} t d t

=∣ \frac{t^{2}}{2} ∣_{0}^{l n 2}

= \frac{{(l n 2)}^{2}}{2}

Commented by prof Abdo imad last updated on 22/May/18

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