Calc II Geometric Series Help. Determing c and r.

[jamesrb]

So I stumbled upon two examples of geometric series. The values given for c and r have me confused.

The first problem:
\(\displaystyle \sum_{n=-4}^{\infty}\left ( -\frac{4}{9} \right )^{n}\)

The answer given has c=1 and r=-4/9. Since the index is not zero its sum is thus:
\(\displaystyle \frac{cr^{-4}}{1-r}=\frac{59,049}{3328}\)

Now the second problem:
\(\displaystyle \sum_{n=1}^{\infty }e^{-n}\)

The answer given has c=1/e and r=1/e also. I can see that the series can be rewritten as:
\(\displaystyle \sum_{n=1}^{\infty }\left ( \frac{1}{e} \right )^{n}\)

Now I get confused because for the sum of this second one they give:
\(\displaystyle \sum_{n=1}^{\infty }e^{n}=\frac{\frac{1}{e}}{1-\frac{1}{e}}\)

I don't see how this follows the formula for the sum of a geometric series starting where the index is not 0: \(\displaystyle \frac{cr^{M}}{1-r}\)

Furthermore I am not sure why c=1/e in the second example and c=1 in the first example. Why doesn't c=1 in the second as it does in the first example. Both of the problems are ratios raised to the n power with no obvious value for c. In fact I don't really see where c is coming from in either example. I can see r easily. I just know that sometimes c turns out to be 1.

 

[HallsofIvy]

So I stumbled upon two examples of geometric series. The values given for c and r have me confused.

The first problem:
\(\displaystyle \sum_{n=-4}^{\infty}\left ( -\frac{4}{9} \right )^{n}\)

The basic formula for geometric sums says that \(\displaystyle \sum{n=0}^\infty cr^n= \frac{c}{1- r}\).
Notice that the "c" term does not depend on n and could be factored out of the sum: \(\displaystyle c\sum_{n=0}^\infty r^n\).

The answer given has c=1 and r=-4/9. Since the index is not zero its sum is thus:
\(\displaystyle \frac{cr^{-4}}{1-r}=\frac{59,049}{3328}\)

This starts at n= -4 rather than 0 so there are two ways to handle this. One is to do the first 4 terms (n= -4, -3, -2, -1) separately: \(\displaystyle (-4/9)^{-4}+ (-4/9)^{-3}+ (-4/9)^{-2}+ (-4/9)^{-1}+ \sum_{n= 0}^\infty \left(-\frac{4}{9}\right)^n\)\(\displaystyle = \frac{6561}{256}- \frac{729}{64}+ \frac{81}{16}- \frac{9}{4}+ \sum_{n=0}^\infty \left(\frac{4}{9}\right)^n\)
The other way way to do this would be to factor out the "extra" terms: \(\displaystyle (-4/9)^-4(-4/9)^4= (-4/9)^0\) so we can write this sum as \(\displaystyle \sum_{n=-4}^\infty (-4/9)^n= \sum_{n=-4}^\infty (-4/9)^n= (-4/9)^{-4}\sum_{n=-4}^\infty (-4/9)^{n+ 4}\) and now let i= n+ 4 so we have \(\displaystyle (-4/9)^{-4}\sum_{i= 0}^\infty (-4/9)^{i}\)
Of course, you could also treat that first \(\displaystyle (-4/9)^4\) as the "c" in the formula.

Now the second problem:
\(\displaystyle \sum_{n=1}^{\infty }e^{n}\)

The answer given has c=1/e and r=1/e also. I can see that the series can be rewritten as:
\(\displaystyle \sum_{n=1}^{\infty }\left ( \frac{1}{e} \right )^{n}\)

??? No, it can't! All you have done is replace \(\displaystyle e^n\) with \(\displaystyle \left(\frac{1}{e}\right)^n\) and those are NOT the same thing at all! Am I misreading this or did you miswrite it?

Again there are two ways to do this:
1) Add the "n= 0" term. \(\displaystyle e^0= 1\), of course. so \(\displaystyle \sum_{n=1}^\infty e^n= \sum_{n=0}^\infty e^{n}- 1\).
2) Factor out "e": \(\displaystyle \sum_{n=1}^\infty= e\sum_{n=0}^\infty e^n\)
This latter is the same as taking c= e and r= e?

Now I get confused because for the sum of this second one they give:
\(\displaystyle \sum_{n=1}^{\infty }e^{n}=\frac{\frac{1}{e}}{1-\frac{1}{e}}\)

I don't see how this follows the formula for the sum of a geometric series starting where the index is not 0: \(\displaystyle \frac{cr^{M}}{1-r}\)

Furthermore I am not sure why c=1/e in the second example and c=1 in the first example. Why doesn't c=1 in the second as it does in the first example. Both of the problems are ratios raised to the n power with no obvious value for c. In fact I don't really see where c is coming from in either example. I can see r easily. I just know that sometimes c turns out to be 1.

I don't understand why you are going from \(\displaystyle e^n\) to \(\displaystyle 1/e^n\).

 

[jamesrb]

The basic formula for geometric sums says that \(\displaystyle \sum{n=0}^\infty cr^n= \frac{c}{1- r}\).
Notice that the "c" term does not depend on n and could be factored out of the sum: \(\displaystyle c\sum_{n=0}^\infty r^n\).


This starts at n= -4 rather than 0 so there are two ways to handle this. One is to do the first 4 terms (n= -4, -3, -2, -1) separately: \(\displaystyle (-4/9)^{-4}+ (-4/9)^{-3}+ (-4/9)^{-2}+ (-4/9)^{-1}+ \sum_{n= 0}^\infty \left(-\frac{4}{9}\right)^n\)\(\displaystyle = \frac{6561}{256}- \frac{729}{64}+ \frac{81}{16}- \frac{9}{4}+ \sum_{n=0}^\infty \left(\frac{4}{9}\right)^n\)
The other way way to do this would be to factor out the "extra" terms: \(\displaystyle (-4/9)^-4(-4/9)^4= (-4/9)^0\) so we can write this sum as \(\displaystyle \sum_{n=-4}^\infty (-4/9)^n= \sum_{n=-4}^\infty (-4/9)^n= (-4/9)^{-4}\sum_{n=-4}^\infty (-4/9)^{n+ 4}\) and now let i= n+ 4 so we have \(\displaystyle (-4/9)^{-4}\sum_{i= 0}^\infty (-4/9)^{i}\)
Of course, you could also treat that first \(\displaystyle (-4/9)^4\) as the "c" in the formula.

??? No, it can't! All you have done is replace \(\displaystyle e^n\) with \(\displaystyle \left(\frac{1}{e}\right)^n\) and those are NOT the same thing at all! Am I misreading this or did you miswrite it?

Again there are two ways to do this:
1) Add the "n= 0" term. \(\displaystyle e^0= 1\), of course. so \(\displaystyle \sum_{n=1}^\infty e^n= \sum_{n=0}^\infty e^{n}- 1\).
2) Factor out "e": \(\displaystyle \sum_{n=1}^\infty= e\sum_{n=0}^\infty e^n\)
This latter is the same as taking c= e and r= e?



I don't understand why you are going from \(\displaystyle e^n\) to \(\displaystyle 1/e^n\).

I have mistyped it, it is supposed to be e^-n. Sorry, one little - sign is hard to find in that great big wall of text. I have edited my original post to include it.