MHB Infinite Geometric Series and Convergence

AI Thread Summary
The discussion focuses on finding the common ratio for an infinite geometric series. For a series with an initial term of 4 that converges to a sum of 16/3, the common ratio is calculated as r = 1/4. Another series is analyzed, leading to a common ratio of r = √5/2, which is greater than 1. Since this ratio exceeds 1, the series diverges. The key takeaway is that a common ratio greater than or equal to 1 indicates divergence in infinite geometric series.
karush
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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?
 
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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?
 
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
 
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$r>1$

Correct. :)
 
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