Question Number 37712 by Rio Mike last updated on 16/Jun/18
$${Evaluate}\:\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}^{{r}−\mathrm{1}} \\ $$
Commented by prakash jain last updated on 17/Jun/18
$$−\mathrm{1}/\mathrm{2}\:\mathrm{is}\:\mathrm{also}\:\mathrm{a}\:\mathrm{valid}\:\mathrm{answer}\:\mathrm{using} \\ $$$$\mathrm{analytical}\:\mathrm{continuity} \\ $$
Answered by Joel579 last updated on 17/Jun/18
$${S}_{{n}} \:=\:\mathrm{2}^{−\mathrm{1}} \:+\:\mathrm{2}^{\mathrm{0}} \:+\:\mathrm{2}^{\mathrm{1}} \:+\:\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{3}} \:+\:\mathrm{2}^{\mathrm{4}} \:+\:…\:+\:\mathrm{2}^{{n}−\mathrm{1}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{S}_{{n}} \:=\:\infty \\ $$
Commented by Rio Mike last updated on 17/Jun/18
$$ \\ $$$$\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}^{{r}−\mathrm{1}} \\ $$$$\mathrm{2}^{\mathrm{0}−\mathrm{1}} +\:\mathrm{2}^{\mathrm{1}−\mathrm{1}} +\mathrm{2}^{\mathrm{2}−\mathrm{1}} +\:\mathrm{2}^{\mathrm{3}−\mathrm{1}} +…\mathrm{2}^{{r}−{n}} \\ $$$${GP}=\:\frac{\mathrm{1}}{\mathrm{2}\:}+\:\mathrm{1}\:+\:\mathrm{2}\:\:+\:\mathrm{4}+… \\ $$$${a}=\mathrm{1}/\mathrm{2}\:\:{and}\:{r}=\:\mathrm{2} \\ $$$${S}_{\infty} =\:\frac{{a}}{\mathrm{1}−{r}} \\ $$$$\:\:\:\:\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\mathrm{2}} \\ $$$$\:\:\:\:=\:−\mathrm{1}/\mathrm{2} \\ $$$$\:\:\:\:\:=\:\:\:\:\:_{} \underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{S}_{{n}} \:=\:\infty\:…… \\ $$
Commented by math1967 last updated on 17/Jun/18
$${For}\:\frac{{a}}{\mathrm{1}−{r}}\:\:{r}<\mathrm{1}\:{then}\:{why}\:{r}=\mathrm{2}\:{applicable}? \\ $$
Commented by Joel579 last updated on 17/Jun/18
$${S}_{\infty} \:=\:\frac{{a}}{\mathrm{1}\:−\:{r}}\:\:\mathrm{only}\:\mathrm{for}\:\mid{r}\mid\:<\:\mathrm{1} \\ $$