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

Evaluate Σ_(r=0) ^∞ 2^(r−1)

$${Evaluate}\:\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}^{{r}−\mathrm{1}} \\ $$

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

−1/2 is also a valid answer using analytical continuity

$$−\mathrm{1}/\mathrm{2}\:\mathrm{is}\:\mathrm{also}\:\mathrm{a}\:\mathrm{valid}\:\mathrm{answer}\:\mathrm{using} \\ $$$$\mathrm{analytical}\:\mathrm{continuity} \\ $$

Answered by Joel579 last updated on 17/Jun/18

S_n = 2^(−1) + 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + ... + 2^(n−1) lim_(n→∞) S_n = ∞

$${S}_{{n}} \:=\:\mathrm{2}^{−\mathrm{1}} \:+\:\mathrm{2}^{\mathrm{0}} \:+\:\mathrm{2}^{\mathrm{1}} \:+\:\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{3}} \:+\:\mathrm{2}^{\mathrm{4}} \:+\:...\:+\:\mathrm{2}^{{n}−\mathrm{1}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{S}_{{n}} \:=\:\infty \\ $$

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

Σ_(r=0) ^∞ 2^(r−1) 2^(0−1) + 2^(1−1) +2^(2−1) + 2^(3−1) +...2^(r−n) GP= (1/(2 ))+ 1 + 2 + 4+... a=1/2 and r= 2 S_∞ = (a/(1−r)) = ((1/2)/(1−2)) = −1/2 = _ lim_(n→0) S_n = ∞ ......

$$ \\ $$$$\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}^{{r}−\mathrm{1}} \\ $$$$\mathrm{2}^{\mathrm{0}−\mathrm{1}} +\:\mathrm{2}^{\mathrm{1}−\mathrm{1}} +\mathrm{2}^{\mathrm{2}−\mathrm{1}} +\:\mathrm{2}^{\mathrm{3}−\mathrm{1}} +...\mathrm{2}^{{r}−{n}} \\ $$$${GP}=\:\frac{\mathrm{1}}{\mathrm{2}\:}+\:\mathrm{1}\:+\:\mathrm{2}\:\:+\:\mathrm{4}+... \\ $$$${a}=\mathrm{1}/\mathrm{2}\:\:{and}\:{r}=\:\mathrm{2} \\ $$$${S}_{\infty} =\:\frac{{a}}{\mathrm{1}−{r}} \\ $$$$\:\:\:\:\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\mathrm{2}} \\ $$$$\:\:\:\:=\:−\mathrm{1}/\mathrm{2} \\ $$$$\:\:\:\:\:=\:\:\:\:\:_{} \underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{S}_{{n}} \:=\:\infty\:...... \\ $$

Commented by math1967 last updated on 17/Jun/18

For (a/(1−r)) r<1 then why r=2 applicable?

$${For}\:\frac{{a}}{\mathrm{1}−{r}}\:\:{r}<\mathrm{1}\:{then}\:{why}\:{r}=\mathrm{2}\:{applicable}? \\ $$

Commented by Joel579 last updated on 17/Jun/18

S_∞ = (a/(1 − r)) only for ∣r∣ < 1

$${S}_{\infty} \:=\:\frac{{a}}{\mathrm{1}\:−\:{r}}\:\:\mathrm{only}\:\mathrm{for}\:\mid{r}\mid\:<\:\mathrm{1} \\ $$

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