Limit Ratio Test: Solving $$\sum_{n=1}^{\infty}\frac{1}{2^n}$$

[Jbreezy]

Homework Statement

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

Homework Equations

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


<br /> \displaystyle\lim_{n\rightarrow \infty} {\frac{2^n}{2^{n+1}}}<br />


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.

 

[Goa'uld]

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

 

[Mark44]

Homework Statement

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

Homework Equations

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 <br /> \displaystyle\lim_{n\rightarrow \infty} {\frac{2^n}{2^{n+1}}}<br />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.

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

 

[Jbreezy]

GRRRRRR! Thanks.

 

[HallsofIvy]

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.

 

[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. ?

 

[micromass]

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. ?

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.

 

[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?

 

[micromass]

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?

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}

 

[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 !

 

[micromass]

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