$\sum _{i=1}^{\mathrm{\infty}}(0.5{)}^{i+1}(i+1)=2$

Yaretzi Mcconnell
2022-11-20
Answered

Could you please explain how come:

$\sum _{i=1}^{\mathrm{\infty}}(0.5{)}^{i+1}(i+1)=2$

$\sum _{i=1}^{\mathrm{\infty}}(0.5{)}^{i+1}(i+1)=2$

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Kennedy Evans

Answered 2022-11-21
Author has **16** answers

We know for $|r|<1$

$\sum _{1\le i<\mathrm{\infty}}{r}^{i+1}={r}^{2}\cdot \frac{1}{1-r}=\frac{{r}^{2}}{1-r}$

$\text{Now,}\frac{d\left(\sum _{1\le i\mathrm{\infty}}{r}^{i+1}\right)}{dr}=\sum _{1\le i\mathrm{\infty}}\frac{d{r}^{i+1}}{dr}=\sum _{1\le i\mathrm{\infty}}(i+1){r}^{i}$

$\text{and}\frac{d\left(\frac{{r}^{2}}{1-r}\right)}{dr}=\frac{2r-{r}^{2}}{(1-r{)}^{2}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\sum _{1\le i<\mathrm{\infty}}(i+1){r}^{i}=\frac{2r-{r}^{2}}{(1-r{)}^{2}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\sum _{1\le i<\mathrm{\infty}}(i+1){r}^{i+1}=\frac{r(2r-{r}^{2})}{(1-r{)}^{2}}$

Alternatively, using Arithmetico-geometric series, we can show,

$\sum _{1\le i<\mathrm{\infty}}(i+c){r}^{i+1}=\frac{(c+1){r}^{2}-c\cdot {r}^{3}}{(1-r{)}^{2}}$

where c is any finite number

$\sum _{1\le i<\mathrm{\infty}}{r}^{i+1}={r}^{2}\cdot \frac{1}{1-r}=\frac{{r}^{2}}{1-r}$

$\text{Now,}\frac{d\left(\sum _{1\le i\mathrm{\infty}}{r}^{i+1}\right)}{dr}=\sum _{1\le i\mathrm{\infty}}\frac{d{r}^{i+1}}{dr}=\sum _{1\le i\mathrm{\infty}}(i+1){r}^{i}$

$\text{and}\frac{d\left(\frac{{r}^{2}}{1-r}\right)}{dr}=\frac{2r-{r}^{2}}{(1-r{)}^{2}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\sum _{1\le i<\mathrm{\infty}}(i+1){r}^{i}=\frac{2r-{r}^{2}}{(1-r{)}^{2}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\sum _{1\le i<\mathrm{\infty}}(i+1){r}^{i+1}=\frac{r(2r-{r}^{2})}{(1-r{)}^{2}}$

Alternatively, using Arithmetico-geometric series, we can show,

$\sum _{1\le i<\mathrm{\infty}}(i+c){r}^{i+1}=\frac{(c+1){r}^{2}-c\cdot {r}^{3}}{(1-r{)}^{2}}$

where c is any finite number

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