show that the close form of Σ_(k=0) ^∞ (Σ_(i=0) ^∞ [(((−1)^(k+i) )/((k+1)(k+2i+2)))])=(1/8)ln2


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Question Number 111622 by mathdave last updated on 04/Sep/20

$s h o w t h a t t h e c l o s e f o r m o f$ $\underset{k = 0}{\sum^{\infty}} (\underset{i = 0}{\sum^{\infty}} [\frac{{(- 1)}^{k + i}}{(k + 1) (k + 2 i + 2)}]) = \frac{1}{8} \ln 2$
	Answered by mathdave last updated on 04/Sep/20

	$s o l u t i o n$ $I = \underset{k = 0}{\sum^{\infty}} (\underset{i = 0}{\sum^{\infty}} [\frac{{(- 1)}^{i + k}}{2 i + 1} (\frac{1}{k + 1} - \frac{1}{k + 2 i + 2})])$ $I = \underset{k = 0}{\sum^{\infty}} (\underset{i = 0}{\sum^{\infty}} [\frac{{(- 1)}^{i + k}}{(2 i + 1) (k + 1)} - \frac{{(- 1)}^{i + k}}{(2 i + 1) (k + 2 i + 2)}])$ $I = (\underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k}}{k + 1}) (\underset{i = 0}{\sum^{\infty}} \frac{{(- 1)}^{i}}{2 i + 1}) - \underset{k = 0}{\sum^{\infty}} \underset{i = 0}{\sum^{\infty}} \frac{{(- 1)}^{i + k}}{(2 i + 1) (k + 2 i + 2)}$ $I = (\underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} \int_{0}^{1} y^{k} d y) (\underset{i = 0}{\sum^{\infty}} {(- 1)}^{i} \int_{0}^{1} m^{2 i} d m) - (\underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} \int_{0}^{1} u^{(k + 2 i + 1)} d u) \underset{i = 0}{\sum^{\infty}} \frac{{(- 1)}^{i}}{2 i + 1}$ $I = (\int_{0}^{1} \frac{d y}{1 + y}) (\int_{0}^{1} \frac{d m}{1 + m^{2}}) - (\underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} \int_{0}^{1} u^{k} d u) \underset{i = 0}{\sum^{\infty}} \frac{{(- 1)}^{i} u^{2 i + 1}}{2 i + 1}$ $b u t \tan^{- 1} (u) = \underset{i = 0}{\sum^{\infty}} \frac{{(- 1)}^{i} u^{2 i + 1}}{2 i + 1}$ $I = (\ln 2) {(\tan^{- 1} (m))}_{0}^{1} - (\int_{0}^{1} \frac{d u}{1 + u}) \tan^{- 1} (u)$ $I = (\ln 2) (\frac{π}{4}) - (\int_{0}^{1} \frac{\tan^{- 1} u}{1 + u} d u)$ $u s i n g I B P$ $I = \frac{π}{4} \ln 2 - [{(\ln (1 + u) \tan^{- 1} u]}_{0}^{1} + (\int_{0}^{1} \frac{\ln (1 + u)}{1 + u^{2}} d u)$ $I = \frac{π}{4} \ln 2 - \frac{π}{4} \ln 2 + \int_{0}^{1} \frac{\ln (1 + u)}{1 + u^{2}} d u$ $I = \int_{0}^{1} \frac{\ln (1 + u)}{1 + u^{2}} d u (l e t u = \tan x a n d d u = (1 + \tan^{2} x) d x$ $I = \int_{0}^{\frac{π}{4}} \ln (1 + \tan x) d x . . . . . . . . . (1) u s i n g t h e p r o p e r t y$ $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ $I = \int_{0}^{\frac{π}{4}} \ln [1 + \tan (\frac{π}{4} - x)] d x =$ $I = \int_{0}^{\frac{π}{4}} \ln (1 + \frac{1 - \tan x}{1 + \tan x}) d x$ $I = \int_{0}^{\frac{π}{4}} (\ln 2 - \ln (1 + \tan x)) d x . . . . . . . . (2)$ $a d d i n g (1) a n d (2)$ $2 I = \int_{0}^{\frac{π}{4}} (\ln (1 + \tan x) + \ln 2 - \ln (1 + \tan x)) d x$ $I = \frac{1}{2} \int_{0}^{\frac{π}{4}} \ln 2 d x = \frac{1}{2} \ln 2 \int_{0}^{\frac{π}{4}} d x = \frac{\ln 2}{2} {[x]}_{0}^{\frac{π}{4}}$ $I = \frac{1}{2} \ln 2 \times \frac{π}{4} = \frac{1}{8} \ln 2$ $∵ \underset{k = 0}{\sum^{\infty}} (\underset{i = 0}{\sum^{\infty}} [\frac{{(- 1)}^{i + k}}{(k + 1) (k + 2 i + 2)}]) = \frac{1}{8} \ln 2 Q . E . D$ $b y m a t h d a v e (04 / 09 / 2020)$
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
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