Infinite Geometric Series and Convergence

[karush]

a. Find the common ration $r$, for an infinite series
with an initial term $4$ that converges to a sum of $\displaystyle\frac{16}{3}$
$$\displaystyle S=\frac{a}{1-r} $$ so $\displaystyle\frac{16}{3}=\frac{4}{1-r}$ then $\displaystyle r=\frac{1}{4}$

b. Consider the infinite geometric series

$\displaystyle
\frac{8}{25} + \frac{4\sqrt{5}}{25}+\frac{2}{5}+\frac{\sqrt{5}}{5}+\cdot\cdot\cdot$

c. What is the exact value of the common ratio of the series
$\frac{8}{25}\cdot r = \frac{4\sqrt{5}}{25}$ so $r=\frac{\sqrt{5}}{2}$d. Does the series converge?

not sure how to do this?

 

[Greg]

Hi karush,

Your final answer for part a is correct.
To find the common ratio, $r$, solve $\dfrac{16}{3}=\dfrac{4}{1-r}$ for $r$.
Please show your work.

Part d:

Is $\dfrac{\sqrt5}{2}$ greater than, less than or equal to $1$? What does your answer to this question imply?

 

[karush]

a. Find the common ration $r$, for an infinite series
with an initial term $4$ that converges to a sum of $\displaystyle\frac{16}{3}$
$$\displaystyle S=\frac{a}{1-r} =$$
so $\displaystyle\frac{16}{3}=\frac{4}{1-r} \\
3\cdot4 = 16\left(1-r\right)\implies12=16-16r\implies4=16r\implies r=\frac{4}{16}=\frac{1}{4}$b. Consider the infinite geometric series

$\displaystyle
\frac{8}{25} + \frac{4\sqrt{5}}{25}+\frac{2}{5}+\frac{\sqrt{5}}{5}+\cdot\cdot\cdot$

c. What is the exact value of the common ratio of the series
$\frac{8}{25}\cdot r = \frac{4\sqrt{5}}{25}$ so $r=\frac{\sqrt{5}}{2}$d. Does the series converge?

$\frac{\sqrt{5}}{2}=1.11$ $r\ge1$ no it divergess

 

[Greg]

$r>1$

Correct. :)