Determinate-lim-n-k-1-n-1-k-1-k-

Question Number 172309 by mathocean1 last updated on 25/Jun/22

D e t e r m i n a t e

\underset{n \to + \infty}{l i m} (\sum_{k = 1}^{n} \frac{{(- 1)}^{k + 1}}{k})

Commented by mr W last updated on 25/Jun/22

\ln (1 + x) = \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k + 1} x^{k}}{k}

w i t h x = 1 :

\ln 2 = \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k + 1}}{k}

Answered by Mathspace last updated on 25/Jun/22

u_{n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k}

a n d p (x) = \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{k} x^{k}

p^{^{'}} (x) = \sum_{k = 1}^{n} {(- 1)}^{k - 1} x^{k - 1}

= \sum_{k = 1}^{n} {(- x)}^{k - 1} = \sum_{k = 0}^{n - 1} {(- x)}^{k}

= \frac{1 - {(- x)}^{n}}{1 + x} \Rightarrow

p (x) = \int_{0}^{x} \frac{1 - {(- t)}^{n}}{1 + t} + c

c = p (0) = 0 \Rightarrow

p (x) = \int_{0}^{x} \frac{1 - {(- t)}^{n}}{1 + t} d t

\Rightarrow p (1) = \int_{0}^{1} \frac{d t}{1 + t} - {(- 1)}^{n} \int_{0}^{1} \frac{t^{n}}{1 + t} d t

{l i m}_{n \to + \infty} u_{n} = {l i m}_{n \to + \infty} p (1)

= l n 2 - {l i m}_{n \to + \infty} {(- 1)}^{n} \int_{0}^{1} \frac{t^{n}}{1 + t} d t

= l n 2 - 0 = l n 2