Is-the-following-correct-S-i-1-2-i-1-S-1-2-4-8-16-32-64-128-2S-2-4-8-6-32-2S-S-1-S-1-

Question Number 4365 by Filup last updated on 13/Jan/16

Is the following correct ?

S = \underset{i = 1}{\sum^{\infty}} 2^{i - 1}

S = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + \dots

∴ 2 S = 2 + 4 + 8 + 6 + 32 + \dots

2 S = S - 1

S = - 1

Commented by Filup last updated on 13/Jan/16

o r

S = \underset{i = n}{\sum^{\infty}} 2^{i}

S = 2^{n} + 2^{n + 1} + 2^{n + 2} + \dots

2 S = 2^{n + 1} + 2^{n + 2} + 2^{n + 3} + \dots

2 S = S - 2^{n}

∴ S = - 2^{n}

c a n b e t a k e n f u r t h e r :

S = \underset{i = n}{\sum^{\infty}} a^{i}

S = a^{n} + a^{n + 1} + a^{n + 2} + \dots

a S = a^{n + 1} + a^{n + 2} + a^{n + 3} + \dots

a S = S - a^{n}

S = - \frac{a^{n}}{a - 1}

∴ S = \frac{a^{n}}{1 - a}

Commented by Filup last updated on 13/Jan/16

is S = \frac{a^{n}}{1 - a} only true as a limiting sum ?

or is it an analytical solution ?

Commented by Yozzii last updated on 13/Jan/16

S (x) = \underset{i = 1}{\sum^{\infty}} x^{i - 1} = 1 + x + x^{2} + x^{3} + \dots

x S (x) = x + x^{2} + x^{3} + x^{4} + \dots

∴ (1 - x) S (x) = 1

S (x) = \frac{1}{1 - x} . T h i s a s s u m e s x \neq 1

a n d t h a t S (x), b e i n g t h e l i m i t o f \underset{i = 1}{\sum^{\infty}} x^{i - 1},

e x i s t s .

N o w, l e t S (x, n) = \underset{i = 1}{\sum^{n}} x^{i - 1} = \frac{x^{n} - 1}{x - 1} (x \neq 1)

S (x, n) = \frac{x^{n}}{x - 1} + \frac{1}{1 - x} .

I f ∣ x ∣> 1 \Rightarrow \lim_{n \to \infty} x^{n} d o e s n o t e x i s t .

\Rightarrow \lim_{n \to \infty} S (x, n) d o e s n o t e x i s t . S o, i f x = 2 > 1

S d o e s n o t h a v e a l i m i t s o t h a t S \neq - 1 .

{I t}^{'} s i n t e r e s t i n g h o w {y o u}^{'} v e s h o w n

S = - 1 . B u t b y t h e c o n c e p t o f n u m b e r

a s a v e c t o r h o w c a n t h e i n f i n i t e

s u m o f p o s i t i v e n u m b e r s (v e c t o r s)

g i v e a n e g a t i v e n u m b e r (o p p o s i t e v e c t o r

d i r e c t i o n) ?

Commented by prakash jain last updated on 13/Jan/16

S = - 1 using analytical continuity .

f (x) = \frac{1}{1 - x}

Commented by malwaan last updated on 14/Jan/16

I t h i n k s = - 1 i s w r o n g e

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