Evaluate-r-0-2-r-1-

Question Number 37712 by Rio Mike last updated on 16/Jun/18

E v a l u a t e \underset{r = 0}{\sum^{\infty}} 2^{r - 1}

Commented by prakash jain last updated on 17/Jun/18

- 1 / 2 is also a valid answer using

analytical continuity

Answered by Joel579 last updated on 17/Jun/18

S_{n} = 2^{- 1} + 2^{0} + 2^{1} + 2^{2} + 2^{3} + 2^{4} + \dots + 2^{n - 1}

\lim_{n \to \infty} S_{n} = \infty

Commented by Rio Mike last updated on 17/Jun/18

\underset{r = 0}{\sum^{\infty}} 2^{r - 1}

2^{0 - 1} + 2^{1 - 1} + 2^{2 - 1} + 2^{3 - 1} + \dots 2^{r - n}

G P = \frac{1}{2} + 1 + 2 + 4 + \dots

a = 1 / 2 a n d r = 2

S_{\infty} = \frac{a}{1 - r}

= \frac{\frac{1}{2}}{1 - 2}

= - 1 / 2

=_{} \lim_{n \to 0} S_{n} = \infty \dots \dots

Commented by math1967 last updated on 17/Jun/18

F o r \frac{a}{1 - r} r < 1 t h e n w h y r = 2 a p p l i c a b l e ?

Commented by Joel579 last updated on 17/Jun/18

S_{\infty} = \frac{a}{1 - r} only for ∣ r ∣ < 1

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