Find the sum of the series.

$\sum _{n=1}^{\mathrm{\infty}}\frac{7}{{4}^{n}}$

$\sum _{n=1}^{\mathrm{\infty}}\frac{7}{{4}^{n}}$

vangstosiis
2022-07-29
Answered

Find the sum of the series.

$\sum _{n=1}^{\mathrm{\infty}}\frac{7}{{4}^{n}}$

$\sum _{n=1}^{\mathrm{\infty}}\frac{7}{{4}^{n}}$

You can still ask an expert for help

kamphundg4

Answered 2022-07-30
Author has **20** answers

$\sum _{n=1}^{\mathrm{\infty}}=7(\frac{1}{4}+(\frac{1}{4}{)}^{2}+(\frac{1}{4}{)}^{3}+...+(\frac{1}{4}{)}^{N}+...)$

$=\frac{7}{4}(1+(\frac{1}{4}{)}^{2}+(\frac{1}{4}{)}^{3}+....)$ this is a geometric series

$r=\frac{1}{4}<1$

and

$1+{r}^{2}+{r}^{3}+{r}^{4}+...........+{r}^{n-1}=\underset{n\to \mathrm{\infty}}{lim}\frac{1-{r}^{n}}{1-r}=\frac{1}{1-r}\text{}if\text{}|r|1$

$\underset{n\to \mathrm{\infty}}{lim}\frac{7}{4}(\frac{1-(\frac{1}{4}{)}^{n}}{1-\frac{1}{4}})=\underset{n\to \mathrm{\infty}}{lim}\frac{7}{4}\ast \frac{1}{\frac{3}{4}}=\frac{7.4}{4.3}=\frac{7}{3}$

$=\frac{7}{4}(1+(\frac{1}{4}{)}^{2}+(\frac{1}{4}{)}^{3}+....)$ this is a geometric series

$r=\frac{1}{4}<1$

and

$1+{r}^{2}+{r}^{3}+{r}^{4}+...........+{r}^{n-1}=\underset{n\to \mathrm{\infty}}{lim}\frac{1-{r}^{n}}{1-r}=\frac{1}{1-r}\text{}if\text{}|r|1$

$\underset{n\to \mathrm{\infty}}{lim}\frac{7}{4}(\frac{1-(\frac{1}{4}{)}^{n}}{1-\frac{1}{4}})=\underset{n\to \mathrm{\infty}}{lim}\frac{7}{4}\ast \frac{1}{\frac{3}{4}}=\frac{7.4}{4.3}=\frac{7}{3}$

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Intermediate value theorem: show there exists $c\in [0,2/3]$ such that $f(c+\frac{1}{3})=f(c)$

Let $f:[0,1]\to [0,1]$ be continuous and $f(0)=f(1)$

This is pretty straight forward and the other ones ive done it was easier to find why IVT applies, namely find why it changes signs.

Letting $g(x)=f(x+\frac{1}{3})-f(x)$ then this is only defined for the interval $[0,\frac{2}{3}]$ b/c not sure what $g(1)=f(\frac{4}{3})-f(1)$ may be.

Now, $g(0)=f(\frac{1}{3})-f(0)$

and $g(\frac{2}{3})=f(1)-f(\frac{2}{3})=f(0)-f(\frac{2}{3})$.

I can't (at least I don't see why I can) conclude this is equal to $-g(0)$ which has been the case in other exercises, which is what I used to apply the IVT and get the wanted result.

Let $f:[0,1]\to [0,1]$ be continuous and $f(0)=f(1)$

This is pretty straight forward and the other ones ive done it was easier to find why IVT applies, namely find why it changes signs.

Letting $g(x)=f(x+\frac{1}{3})-f(x)$ then this is only defined for the interval $[0,\frac{2}{3}]$ b/c not sure what $g(1)=f(\frac{4}{3})-f(1)$ may be.

Now, $g(0)=f(\frac{1}{3})-f(0)$

and $g(\frac{2}{3})=f(1)-f(\frac{2}{3})=f(0)-f(\frac{2}{3})$.

I can't (at least I don't see why I can) conclude this is equal to $-g(0)$ which has been the case in other exercises, which is what I used to apply the IVT and get the wanted result.