What is the Limit of a Geometric Series with a Fractional Common Ratio?

[togame]

Homework Statement

Evaluate the problem: $$\sum_{k=1}^\infty \frac{3^{(k-1)}}{4^{(k+1)}}$$

Homework Equations

$\displaystyle\lim_{n\rightarrow\infty} S_n = \frac{a}{1-r}$

The Attempt at a Solution

I know that the limit of the partial sequence is what i need to help solve this, but can't figure out how to do that. Can anyone point me in the right direction?

 

[algebrat]

Homework Statement

Evaluate the problem: $$\sum_{k=1}^\infty \frac{3^{(k-1)}}{4^{(k+1)}}$$

Homework Equations

$\displaystyle\lim_{n\rightarrow\infty} S_n = \frac{a}{1-r}$

The Attempt at a Solution

I know that the limit of the partial sequence is what i need to help solve this, but can't figure out how to do that. Can anyone point me in the right direction?

Factor some terms out so it looks like (3/4)^k, and k=0..∞

EDIT: cleanest way might be pullout 1/16 and shift your index, an alternative might be pulling out 1/12 and adding (and subtracting) 1 (or 1/12 rather).

SECOND EDIT: I apologize, I didn't read your question carefully.

 

[Ray Vickson]

Homework Statement

Evaluate the problem: $$\sum_{k=1}^\infty \frac{3^{(k-1)}}{4^{(k+1)}}$$

Homework Equations

$\displaystyle\lim_{n\rightarrow\infty} S_n = \frac{a}{1-r}$

The Attempt at a Solution

I know that the limit of the partial sequence is what i need to help solve this, but can't figure out how to do that. Can anyone point me in the right direction?

Start by figuring out what is r and what is a. Also, in $S_n,$ what values of n are included (i.e., where does n "start")?

RGV

 

[HallsofIvy]

Start by figuring out what is r and what is a. Also, in $S_n,$ what values of n are included (i.e., where does n "start")?

RGV

Related to Ray Vickson's last point, the sum of a geometric series is
$$\sum_{n=0}^\infty ar^n= \frac{a}{1- r}$$

Your sum is
$$\sum_{n= 1}^\infty \frac{3^n}{4^n}= \sum_{n=1}^\infty \left(\frac{3}{4}\right)^n$$

Do you see the difference?

 

[Saitama]

Related to Ray Vickson's last point, the sum of a geometric series is
$$\sum_{n=0}^\infty ar^n= \frac{a}{1- r}$$

Your sum is
$$\sum_{n= 1}^\infty \frac{3^n}{4^n}= \sum_{n=1}^\infty \left(\frac{3}{4}\right)^n$$

Do you see the difference?

Don't forget the 1/12 factor taken out in the HOI equation.

 

[algebrat]

Just as $(1-r)(1+r)=1-r^2$, we have $(1-r)(1+r+r^2+\cdots+r^n)=1-r^{n+1}$. So if $r\ne 1$, then
$$1+r+r^2+\cdots+r^n=\frac{1-r^{n+1}}{1-r}.$$
Additionally, if $r<1$, then we can take the limit of both sides. Notice the left side is your partial sum. Also, you may multiply both sides by $a$.