# Math Help - [SOLVED] Bit of Confusion with the geometric series

1. ## [SOLVED] Bit of Confusion with the geometric series

I see the geometric series stated in two ways, in text books and other sources on the internet.

Generally an infinite geometric series has the form

$a+ax+ax^2+.....ax^{n-1}+ax^n$

and a finite form of the geometric series generally is as such

$a+ax+ax^2+.....ax^{n-2}+ax^{n-1}$

Now for example, even thought I've found my own way of solving this problem, when I come to others that involve taking a finite form of the partial sum, I am unsure if my method is good enough, even though it is correct 90% of the time, I want to be accurate.

Example:

Here is the infinite series of
$\sum_{n=4}^{\infty}(\frac{1}{3})^n$

Now here is the finite form of the series of to the 20th term

$\sum_{n=4}^{20}(\frac{1}{3})^n$

The finite form of the summation was written as such, but they state the general form of a finitie series is to the n-1 power. So I would rewrite this as, if I followed the form that was mentioned

$\sum_{n=4}^{20}(\frac{1}{3})^{n-1}$

Now If I follow this, the first term I get is $(\frac{1}{3})^3$

Following the form for a finite series partial sum is

$\frac{\frac{1}{3}^3(1-\frac{1}{3}^{20})}{1-\frac{1}{3}}$
However, this is not quite right as provided by the book or other solutions I've seen my friends do of this problem, when I do the way I found is more accurate for me

$\sum_{n=4}^{20}(\frac{1}{3})^n$

Where I look at the finite series partial sum form as this, the first term unbroken to help simplifying exponents

$\frac{ar^n-ar^{n+1}}{1-r}$

Where $ar^n$ is the first term in the series and $ar^{n+1}$ is the last term in the series + 1, works the same for the general case of the nth term

$\frac{(\frac{1}{3}^4-\frac{1}{3}^{21})}{1-\frac{1}{3}}$

Similarily

$\frac{\frac{1}{3}^4(1-\frac{1}{3}^{20})}{1-\frac{1}{3}}$

Which is the correct term before further simplifiction, so whats the deal with the form they mention and other sources I have seen explaining the geometric series? Confuses me at times. Sorry for the lengthy post, I know how MHF want poster to include all information

2. Originally Posted by RockHard
I see the geometric series stated in two ways, in text books and other sources on the internet.

Generally an infinite geometric series has the form

$a+ax+ax^2+.....ax^{n-1}+ax^n$

and a finite form of the geometric series generally is as such

$a+ax+ax^2+.....ax^{n-2}+ax^{n-1}$

Now for example, even thought I've found my own way of solving this problem, when I come to others that involve taking a finite form of the partial sum, I am unsure if my method is good enough, even though it is correct 90% of the time, I want to be accurate.

Example:

Here is the infinite series of
$\sum_{n=4}^{\infty}(\frac{1}{3})^n$

Now here is the finite form of the series of to the 20th term

$\sum_{n=4}^{20}(\frac{1}{3})^n$

The finite form of the summation was written as such, but they state the general form of a finitie series is to the n-1 power. So I would rewrite this as, if I followed the form that was mentioned

$\sum_{n=4}^{20}(\frac{1}{3})^{n-1}$

Now If I follow this, the first term I get is $(\frac{1}{3})^3$

Following the form for a finite series partial sum is

$\frac{\frac{1}{3}^3(1-\frac{1}{3}^{20})}{1-\frac{1}{3}}$
However, this is not quite right as provided by the book or other solutions I've seen my friends do of this problem, when I do the way I found is more accurate for me

$\sum_{n=4}^{20}(\frac{1}{3})^n$

Where I look at the finite series partial sum form as this, the first term unbroken to help simplifying exponents

$\frac{ar^n-ar^{n+1}}{1-r}$

Where $ar^n$ is the first term in the series and $ar^{n+1}$ is the last term in the series + 1, works the same for the general case of the nth term

$\frac{(\frac{1}{3}^4-\frac{1}{3}^{21})}{1-\frac{1}{3}}$

Similarily

$\frac{\frac{1}{3}^4(1-\frac{1}{3}^{20})}{1-\frac{1}{3}}$

Which is the correct term before further simplifiction, so whats the deal with the form they mention and other sources I have seen explaining the geometric series? Confuses me at times. Sorry for the lengthy post, I know how MHF want poster to include all information
I think the $n-1$ comes from the sum of the first $n$ terms (summing from $k=0$ to $k=n-1$).

$\underbrace{a+ax+ax^2+...+ax^{n-1}}_{n~terms}$

For the two sums you mentioned,

$\sum_{n=4}^{20}\frac{1}{3^n}=\frac{\frac{1}{3^4}-\frac{1}{3^{21}}}{1-\frac{1}{3}}$

$\sum_{n=4}^{20}\frac{1}{3^{n-1}}=\sum_{n=3}^{19}\frac{1}{3^n}=\frac{\frac{1}{3^ 3}-\frac{1}{3^{20}}}{1-\frac{1}{3}}$

They're different sums. How you choose to write a given sum is up to you.

3. Ah, Thank you for some clarification. Also I tried this

$(\frac{1}{3})^4\sum_{n=0}^{\infty}(\frac{1}{3})^n$

Yielded the same result of $\frac{(\frac{1}{3}^4)(1-\frac({1}{3})^{17})}{1-\frac{1}{3}}$

4. Originally Posted by RockHard
I see the geometric series stated in two ways, in text books and other sources on the internet.

Generally an infinite geometric series has the form

$a+ax+ax^2+.....ax^{n-1}+ax^n$
Well, no. An infinite series does not end:
$a+ ax+ ax^2+ ....$.

and a finite form of the geometric series generally is as such

$a+ax+ax^2+.....ax^{n-2}+ax^{n-1}$
This is exactly the same as you wrote above, just with a different "n".

Now for example, even thought I've found my own way of solving this problem, when I come to others that involve taking a finite form of the partial sum, I am unsure if my method is good enough, even though it is correct 90% of the time, I want to be accurate.

Example:

Here is the infinite series of
$\sum_{n=4}^{\infty}(\frac{1}{3})^n$
You think of this in two different ways. The first term is with n= 4 so the first term is $\frac{1}{3^4}= \frac{1}{81}$.

One thing you could do is factor out \frac{1}{81}[/tex]:
$\frac{1}{81}\sum_{n=0}^\infty \left(\frac{1}{3}\right)^n$
Now, instead of have part of a series with a= 1, r= 1/3, you have the entire series with a= 1/81, r= 1/3.

The other way is to "add and subtract" the missing n=1, n= 2, and n= 3 terms:
$\sum_{n=4}^\infty \left(\frac{1}{3}\right)^n= \sum_{n=0}^\infty \left(\frac{1}{3}\right)^n- 1- \frac{1}{9}- \frac{1}{27}$.

[quote]Now here is the finite form of the series of to the 20th term

$\sum_{n=4}^{20}(\frac{1}{3})^n$
Again, you could factor out 1/81 to get [tex]\frac{1}{81}\sum_{n=4}^{20}\left{1}{3}\right)^{n-4}
and now, if you want that to be something like "n-1" power, change the number so that n-4= m- 1: that means m= n-4+1= n- 3. When n= 4, m= 1 and when n= 20, m= 17. The series becomes $\frac{1}{81}\sum_{m=1}^17 \left(\frac{1}{3}\right)^{m-1}$.

The finite form of the summation was written as such, but they state the general form of a finitie series is to the n-1 power. So I would rewrite this as, if I followed the form that was mentioned

$\sum_{n=4}^{20}(\frac{1}{3})^{n-1}$

Now If I follow this, the first term I get is $(\frac{1}{3})^3$

Following the form for a finite series partial sum is

$\frac{\frac{1}{3}^3(1-\frac{1}{3}^{20})}{1-\frac{1}{3}}$
However, this is not quite right as provided by the book or other solutions I've seen my friends do of this problem, when I do the way I found is more accurate for me

$\sum_{n=4}^{20}(\frac{1}{3})^n$

Where I look at the finite series partial sum form as this, the first term unbroken to help simplifying exponents

$\frac{ar^n-ar^{n+1}}{1-r}$

Where $ar^n$ is the first term in the series and $ar^{n+1}$ is the last term in the series + 1, works the same for the general case of the nth term

$\frac{(\frac{1}{3}^4-\frac{1}{3}^{21})}{1-\frac{1}{3}}$

Similarily

$\frac{\frac{1}{3}^4(1-\frac{1}{3}^{20})}{1-\frac{1}{3}}$

Which is the correct term before further simplifiction, so whats the deal with the form they mention and other sources I have seen explaining the geometric series? Confuses me at times. Sorry for the lengthy post, I know how MHF want poster to include all information