Geometric Series

Section: 
9.2
Date: 
Friday, October 1, 2010 - 14:00 - 15:00
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fall2010math2300_geometric-series.pdf59.1 KB

Math 2300 Section 005 – Calculus II – Fall 2010

9.2 Geometric Series Notes – October 1, 2010

A tip on recognizing geometric series: Many times when you're using convergence comparison tests, the series you use for comparison is either a p-series or a geometric series. In the first case, we know that

\sum _{{n=1}}^{\infty}\frac{1}{x^{p}}

converges for p>1 and diverges for p\le 1. In the second case, the geometric series in question may be more challenging to pick out. For instance, when you have

\sum _{{n=1}}^{\infty}\frac{1}{2^{n}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots

you may either view this as the geometric series

\sum _{{n=1}}^{\infty}\frac{1}{2}\left(\frac{1}{2}^{{n-1}}\right)

where a=1/2 and x=1/2, giving from the formulas for geometric series that

\sum _{{n=1}}^{\infty}\frac{1}{2^{n}}=\frac{1/2}{1-1/2}=\frac{1/2}{1/2}=1.

Or (and this may seem odd at first, but I think it yields faster computations) view this as the geometric series

1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots

where a=1 and x=1/2, except that the first term needs to be removed (since the series in question has no 1). So, we also have

\sum _{{n=1}}^{\infty}\frac{1}{2^{n}}=\frac{1}{1-1/2}-1=1.

WebWork Problem 9.2 #6 A ball is dropped from a height of 15 feet and bounces. Suppose that each bounce is 7/8 of the height of the bounce before. Thus, after the ball hits the floor for the first time, it rises to a height of 15(\frac{7}{8}) feet. Assume that g=32\,ft/s{}^{2} and no air resistance.

A. Find an expression for the height, in feet, to which the ball rises after it hits the floor for the nth time.

Solution: The ball is dropped from 15 feet. It then bounces 15\left(\frac{7}{8}\right) feet, and then it bounces 15\left(\frac{7}{8}\right)^{2} feet, and then it bounces 15\left(\frac{7}{8}\right)^{3} feet. Thus, after the nth bounce we get a height of

b_{n}=15\left(\frac{7}{8}\right)^{n}\text{ feet}

B. Find an expression for the total vertical distance the ball has traveled, in feet, when it hits the floor for the first, second, third and fourth times.

Solution: The height at the time of the first bounce is just 15 feet. Then, we go up 15\left(\frac{7}{8}\right) feet and come back down 15\left(\frac{7}{8}\right) before we bounce again. So, the total height traveled at the time of the second bounce is

15+2\cdot 15\left(\frac{7}{8}\right)\text{ feet}.

At the time of the third bounce, we have traveled

15+2\cdot 15\left(\frac{7}{8}\right)+2\cdot 15\left(\frac{7}{8}\right)^{2}\text{ feet}

and after the fourth bounce we have traveled

15+2\cdot 15\left(\frac{7}{8}\right)+2\cdot 15\left(\frac{7}{8}\right)^{2}+3\cdot 15\left(\frac{7}{8}\right)^{3}\text{ feet}.

C. Find an expression, in closed form, for the total vertical distance the ball has traveled when it hits the floor for the nth time.

Solution: We would like to create a geometric series from the terms given above. The only problem is that the first term in the series doesn't precisely follow the pattern, since every other term in the series is multiplied by 2. So, what we do is we create the series as if the first term followed the pattern and then subtract by 15. So, the formula for the distance traveled by the ball when it hits the floor for the nth time is

\left[\sum _{{j=1}}^{n}2\cdot 15\left(\frac{7}{8}\right)^{{j-1}}\right]-15=\left[30\frac{1-\left(\frac{7}{8}\right)^{n}}{1-\frac{7}{8}}\right]-15

Proof of the Geometric Series Formula:

We need to use the (slightly obscure) fact that

\frac{1-x^{n}}{1-x}=\frac{x^{n}-1}{x-1}=x^{{n-1}}+x^{{n-2}}+\cdots+x^{2}+x+1.

Thus, if we have a geometric series, we can write

\displaystyle\sum _{{k=1}}^{n}ax^{{k-1}} \displaystyle=a+ax+ax^{2}+\cdots+ax^{{n-1}}
\displaystyle=a(1+x+x^{2}+\cdots+x^{{n-1}})
\displaystyle=a\frac{1-x^{n}}{1-x}.

This formula itself allows us to sum any finite geometric series. Next, we take the limit as n\rightarrow\infty to form the formula for the infinite geometric series.

\lim _{{n\rightarrow\infty}}\sum _{{k=1}}^{n}ax^{{k-1}}=\lim _{{n\rightarrow\infty}}a\frac{1-x^{n}}{1-x}=a\frac{1}{1-x}=\frac{a}{1-x}

provided that |x|<1. If |x|\ge 1 then the series will clearly never converge as we will continue to add more and more stuff of size at least ax.

© 2011 Jason B. Hill. All Rights Reserved.