# Limit ratio test

1. Jul 10, 2013

### Jbreezy

1. The problem statement, all variables and given/known data

$$\sum_{n=1} ^\infty\frac{1} {2^n}$$

2. Relevant equations

3. The attempt at a solution
I know just by looking at it that it converges no problem. You do the ratio test and you get something of the form

$\displaystyle\lim_{n\rightarrow \infty} {\frac{2^n}{2^{n+1}}}$

If $$2^n$$ goes to infinity. And $$2^{n+1}$$ does also. Then how is this equal to 1/2 ?It should be infinity over infinity. Which I know it can't be since I already know it converges by looking.

2. Jul 10, 2013

### Goa'uld

Before you take the limit, simplify $\frac{2^{n}}{2^{n+1}}$.

3. Jul 10, 2013

### Staff: Mentor

Because
$$\lim_{n\to \infty} \frac{2^n}{2^{n+1}}= \lim_{n\to \infty} \frac{2^n}{2 \cdot 2^n} = \frac 1 2$$

4. Jul 10, 2013

### Jbreezy

GRRRRRR!!!!!!!!!!!!!! Thanks.

5. Jul 10, 2013

### HallsofIvy

Staff Emeritus
In fact, except for the fact that the sum starts at n= 1 rather than n= 0, that's a geometric series. It is easy to show that $$\sum_{n=0}^\infty r^n= \frac{1}{1- r}$$. Calculate that with r=1/2 then subtract off the value when n= 0, which is $$(1/2)^0= 1$$.

6. Jul 10, 2013

### Jbreezy

Yeah I saw it was geometric they just wanted me to to use the ratio test. It is funny you mention this. I want to know why does it matter whether you start and n =1 or n =0. ?

7. Jul 10, 2013

### micromass

Staff Emeritus
For checking convergence, it doesn't matter whether you start with 0 or 1.
If you want to find the actual value of the sum, then it does matter because

$$\sum_{n=0}^{+\infty} r^n = 1 + \sum_{n=1}^{+\infty} r^n$$

So the results are different.

8. Jul 10, 2013

### Jbreezy

The original was $\sum_{n=1} ^\infty\frac{1} {2^n}$

Why would you not just do

$\sum_{n=1}^\infty r^n= \frac{1}{1- r}$

notice n starts at 1. Why did Halls of Ivy change it then subtract off the one? I mean I get why but why not just start with the original at n = 1 and sum? Does my question make any sense?

9. Jul 10, 2013

### micromass

Staff Emeritus
Because

$$\sum_{n=1}^\infty r^n= \frac{1}{1- r}$$

is not true. The equality is only true when $n$ starts with $0$.

That said, we do have

$$\sum_{n=1}^\infty r^n= \frac{1}{1- r} = -1+\sum_{n=0}^\infty r^n = -1+\frac{1}{1-r} = \frac{r}{1-r}$$

Last edited: Jul 10, 2013
10. Jul 10, 2013

### Jbreezy

You wrote ..$$\sum_{n=1}^\infty r^n= \frac{1}{1- r} = -1+\sum_{n=1}^\infty r^n = -1+\frac{1}{1-r} = \frac{r}{1-r}$$

Did you mean...

$$\sum_{n=0}^\infty r^n= \frac{1}{1- r} = -1+\sum_{n=1}^\infty r^n = -1+\frac{1}{1-r} = \frac{r}{1-r}$$
? Or I'm mistaken?
Thank you !

11. Jul 10, 2013

### micromass

Staff Emeritus
Neither. I have corrected the post. Thanks for spotting the error!