(1/2) + (3/4) + (5/8) + (7/(16)) + ....... = ?


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Question Number 2941 by Syaka last updated on 30/Nov/15

$\frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \frac{7}{16} + . . . . . . . = ?$
Commented by Syaka last updated on 01/Dec/15

$T h a n k s f o r S o l u t i o n S i r R a s h e e d$ $a n d a l s o f o r S o l v e d f r o m S i r Y o z z i . L i k e t h a t .$
	Answered by Rasheed Soomro last updated on 01/Dec/15

	$L e t S = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \frac{7}{16} + . . . . . . . = ?$ $G e n r a l t e r m$ $= \frac{1 + (n - 1) \times 2}{2^{n}} = \frac{2 n - 1}{2^{n}} = \frac{2 n}{2^{n}} - \frac{1}{2^{n}} = \frac{n}{2^{n - 1}} - \frac{1}{2^{n}}$ $S o,$ $S = \underset{n = 1}{\overset{\infty}{Σ}} (\frac{n}{2^{n - 1}} - \frac{1}{2^{n}}) = \underset{n = 1}{\overset{\infty}{Σ}} (\frac{n}{2^{n - 1}}) - \underset{n = 1}{\overset{\infty}{Σ}} (\frac{1}{2^{n}})$ $S = (\frac{1}{2^{0}} - \frac{1}{2^{1}}) + (\frac{2}{2^{1}} - \frac{1}{2^{2}}) + (\frac{3}{2^{2}} - \frac{1}{2^{3}}) + (\frac{4}{2^{3}} - \frac{1}{2^{4}}) + . . .$ $S = (\frac{1}{2^{0}} + \frac{2}{2^{1}} + \frac{3}{2^{2}} + \frac{4}{2^{3}} + . . .) - (\frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \frac{1}{2^{4}} + . . .)$ $S = (\frac{1}{2^{0}} + \frac{2}{2^{1}} + \frac{3}{2^{2}} + \frac{4}{2^{3}} + . . .) - (\frac{\frac{1}{2}}{1 - \frac{1}{2}})$ $S = (1 + \frac{2}{2^{1}} + \frac{3}{2^{2}} + \frac{4}{2^{3}} + . . .) - 1$ $S = \frac{2}{2^{1}} + \frac{3}{2^{2}} + \frac{4}{2^{3}} + . . . = \frac{1}{2} [\frac{2}{2^{0}} + \frac{3}{2^{1}} + \frac{4}{2^{2}} + . . .]$ $G e n e r a l t e r m : \frac{n + 1}{2^{n - 1}} = \frac{n}{2^{n - 1}} + \frac{1}{2^{n - 1}}$ $S = \frac{1}{2} [(\frac{1}{2^{1}} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + . . .) + (\frac{1}{2^{0}} + \frac{1}{2^{1}} + \frac{1}{2^{2}} + . . .)]$ $S = \frac{1}{2} [(\frac{1}{2^{0}} + \frac{2}{2^{1}} + \frac{3}{2^{2}} + . . .) + \frac{1}{1 - \frac{1}{2}}]$ $S = \frac{1}{2} [(\frac{1}{2^{0}} + \frac{2}{2^{1}} + \frac{3}{2^{2}} + . . .) + 2] > \frac{1}{2} [(\frac{1}{2^{0}} + \frac{1}{2^{1}} + \frac{1}{2^{2}} + . . .) + 2]$ $S > \frac{1}{2} [\frac{1}{1 - \frac{1}{2}} + 2] = \frac{1}{2} (2 + 2) = 2$ $S > 2$ $C o n t i n u e$
	Commented by Yozzi last updated on 01/Dec/15
	$Let Q=Σ_(n=1) ^N nx^(n−1) =Σ_(n=1) ^N (d/dx)(x^n ) (x∈R) Q=(d/dx)(Σ_(n=1) ^N x^n ) Q=(d/dx)(xΣ_(n=1) ^N x^(n−1) ) Q=(d/dx)(x((x^N −1)/(x−1))) (x≠1) Q=(d/dx)(((x^(N+1) −x)/(x−1))) Q=(((x−1)((N+1)x^N −1)−(1)(x^(N+1) −x))/((x−1)^2 )) Q=(((N+1)x^(N+1) −x−(N+1)x^N +1−x^(N+1) +x)/((x−1)^2 )) Q=((Nx^(N+1) −(N+1)x^N +1)/((x−1)^2 )) Let x=0.5 ∴Q=((N(0.5)^(N+1) −(N+1)(0.5)^N +1)/((0.5−1)^2 )) Q=4(N(0.5)^(N+1) −(N+1)(0.5)^N +1) lim_(N→∞) Q=4{lim_(N→∞) N(0.5)^(N+1) −lim_(N→∞) (N+1)(0.5)^N +1} N(0.5)^(N+1) =(N/2^(N+1) )→(∞/∞) (indeterminate) N→∞. By use of L′Hopital′s rule, lim_(N→∞) (N/2^(N+1) )=lim_(N→∞) (1/(2^(N+1) ln2))=(1/∞)=0 and similarly lim_(N→∞) (N+1)(0.5)^N =0 ∴ lim_(N→∞) Q=4(0−0+1)=4 ⇒S=4−((1/2)/(1−0.5))=3$
	$L e t Q = \underset{n = 1}{\sum^{N}} {n x}^{n - 1} = \underset{n = 1}{\sum^{N}} \frac{d}{d x} (x^{n}) (x \in R)$ $Q = \frac{d}{d x} (\underset{n = 1}{\sum^{N}} x^{n})$ $Q = \frac{d}{d x} (x \underset{n = 1}{\sum^{N}} x^{n - 1})$ $Q = \frac{d}{d x} (x \frac{x^{N} - 1}{x - 1}) (x \neq 1)$ $Q = \frac{d}{d x} (\frac{x^{N + 1} - x}{x - 1})$ $Q = \frac{(x - 1) ((N + 1) x^{N} - 1) - (1) (x^{N + 1} - x)}{{(x - 1)}^{2}}$ $Q = \frac{(N + 1) x^{N + 1} - x - (N + 1) x^{N} + 1 - x^{N + 1} + x}{{(x - 1)}^{2}}$ $Q = \frac{{N x}^{N + 1} - (N + 1) x^{N} + 1}{{(x - 1)}^{2}}$ $L e t x = 0.5$ $∴ Q = \frac{N {(0.5)}^{N + 1} - (N + 1) {(0.5)}^{N} + 1}{{(0.5 - 1)}^{2}}$ $Q = 4 (N {(0.5)}^{N + 1} - (N + 1) {(0.5)}^{N} + 1)$ $\lim_{N \to \infty} Q = 4 {\lim_{N \to \infty} N {(0.5)}^{N + 1} - \lim_{N \to \infty} (N + 1) {(0.5)}^{N} + 1}$ $N {(0.5)}^{N + 1} = \frac{N}{2^{N + 1}} \to \frac{\infty}{\infty} (i n d e t e r m i n a t e)$ $N \to \infty . B y u s e o f L^{'} {H o p i t a l}^{'} s r u l e,$ $\lim_{N \to \infty} \frac{N}{2^{N + 1}} = \lim_{N \to \infty} \frac{1}{2^{N + 1} l n 2} = \frac{1}{\infty} = 0 a n d$ $s i m i l a r l y \lim_{N \to \infty} (N + 1) {(0.5)}^{N} = 0$ $∴ \lim_{N \to \infty} Q = 4 (0 - 0 + 1) = 4$ $\Rightarrow S = 4 - \frac{1 / 2}{1 - 0.5} = 3$
	Commented by RasheedAhmad last updated on 02/Dec/15

	$A p p r o a c h t h a t I {c o u l d n}^{'} t e v e n t h i n k o f!$
	Answered by Rasheed Soomro last updated on 03/Dec/15

	$L e t$ $S = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \frac{7}{16} + . . . . . .$ $G e n e r a l T e r m : \frac{2 n - 1}{2^{n}} = \frac{n}{2^{n - 1}} - \frac{1}{2^{n}}$ $S = \underset{n = 1}{\overset{\infty}{Σ}} (\frac{n}{2^{n - 1}} - \frac{1}{2^{n}}) = \underset{n = 1}{\overset{\infty}{Σ}} (\frac{n}{2^{n - 1}}) - \underset{n = 1}{\overset{\infty}{Σ}} (\frac{1}{2^{n}})$ $= \underset{n = 1}{\overset{\infty}{Σ}} (n {(\frac{1}{2})}^{n - 1}) - \frac{\frac{1}{2}}{1 - \frac{1}{2}}$ $= \underset{n = 1}{\overset{\infty}{Σ}} (n {(\frac{1}{2})}^{n - 1}) - 1$ $W i t h t h e h e l p o f Y o z z i^{'} s c o m m e n t$ $L e t \frac{1}{2} (w r i t t e n i n r e d) \to x$ $F o r x = \frac{1}{2} :$ $S = \underset{n = 1}{\overset{\infty}{Σ}} ({n x}^{n - 1}) - 1$ $S = - 1 + \underset{n = 1}{\overset{\infty}{Σ}} \frac{d}{d x} (x^{n})$ $S = - 1 + \frac{d}{d x} (\underset{N \to \infty}{l i m} \underset{n = 1}{\overset{N}{Σ}} (x^{n}))$ $= - 1 + \underset{N \to \infty}{l i m} [\frac{d}{d x} (\frac{x (x^{N} - 1)}{x - 1})]$ $= - 1 + \underset{N \to \infty}{l i m} [\frac{d}{d x} (\frac{x^{N + 1} - 1}{x - 1})]$ $= - 1 + \underset{N \to \infty}{l i m} [\frac{(x - 1) \frac{d}{d x} (x^{N + 1} - 1) - (x^{N + 1} - 1) \frac{d}{d x} (x - 1)}{{(x - 1)}^{2}}]$ $= - 1 + \underset{N \to \infty}{l i m} [\frac{(x - 1) (N + 1) x^{N} - (x^{N + 1} - 1) (1)}{{(x - 1)}^{2}}]$ $= - 1 + \underset{N \to \infty}{l i m} [\frac{(N + 1) x^{N + 1} - (N + 1) x^{N} - x^{N + 1} + 1}{{(x - 1)}^{2}}]$ $= - 1 + \underset{N \to \infty}{l i m} [\frac{(N + 1) x^{N + 1} - x^{N + 1} - (N + 1) x^{N} + 1}{{(x - 1)}^{2}}]$ $= - 1 + \underset{N \to \infty}{l i m} [\frac{(N + 1 - 1) x^{N + 1} - (N + 1) x^{N} + 1}{{(x - 1)}^{2}}]$ $= - 1 + \underset{N \to \infty}{l i m} [\frac{{N x}^{N + 1} - (N + 1) x^{N} + 1}{{(x - 1)}^{2}}]$ $∵ x = \frac{1}{2}$ $= - 1 + \underset{N \to \infty}{l i m} [\frac{N {(\frac{1}{2})}^{N + 1} - (N + 1) {(\frac{1}{2})}^{N} + 1}{{(\frac{1}{2} - 1)}^{2}}]$ $= - 1 + \underset{N \to \infty}{l i m} [\frac{N {(\frac{1}{2})}^{N + 1} - (N + 1) {(\frac{1}{2})}^{N} + 1}{\frac{1}{4}}]$ $= - 1 + 4 \underset{N \to \infty}{l i m} [\frac{N}{2^{N + 1}} - \frac{N + 1}{2^{N}} + 1]$ $= - 1 + 4 [\underset{N \to \infty}{l i m} \frac{N}{2^{N + 1}} - \underset{N \to \infty}{l i m} \frac{N + 1}{2^{N}} + \underset{N \to \infty}{l i m} 1]$ $B y L^{'} H o p i t a l^{'} s r u l e \underset{N \to \infty}{l i m} \frac{N}{2^{N + 1}} = 0, \underset{N \to \infty}{l i m} \frac{N + 1}{2^{N}} = 0$ $= - 1 + 4 (1) = 3$
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