prove-that-n-0-1-n-n-1-ln2-

Question Number 161755 by mkam last updated on 22/Dec/21

p r o v e t h a t \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + 1} = l n 2

Answered by FelipeLz last updated on 22/Dec/21

f (x) = \ln (x + 1)

f^{'} (x) = \frac{1}{x + 1}

f^{″} (x) = - \frac{1}{{(x + 1)}^{2}}

f^{‴} (x) = \frac{2}{{(x + 1)}^{3}}

f^{⁗} (x) = - \frac{6}{{(x + 1)}^{4}}

⋮

f^{(k)} (x) = {(- 1)}^{k - 1} \frac{(k - 1)!}{{(x + 1)}^{k}}

f (x) = \underset{k = 0}{\sum^{\infty}} \frac{f^{(k)} (a)}{k!} {(x - a)}^{k}

a = 0 \Rightarrow {\begin{cases} f (a) = 0 \\ f^{(k)} (a) = {(- 1)}^{k - 1} (k - 1)! \end{cases} ∴ f (x) = \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k - 1} (k - 1)!}{k!} x^{k} = \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k - 1} x^{k}}{k}

k = n + 1 \Rightarrow f (x) = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{n + 1}}{n + 1}

\ln (2) = f (1) = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} {(1)}^{n + 1}}{n + 1} = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + 1}

Answered by Ar Brandon last updated on 22/Dec/21

S = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + 1} = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} x^{n} d x

= \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} {(- x)}^{n} d x = \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} {(- x)}^{n} d x

= \int_{0}^{1} \frac{1}{1 + x} d x = {[\ln (1 + x)]}_{0}^{1} = \ln (2) - \ln (1)

\Rightarrow \begin{matrix} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + 1} = \ln (2) \end{matrix}