# Math Help - Infinite Series

1. ## Infinite Series

How do I prove: $\sum_{n=1}^{\infty} \frac {(-1)^n} {2^n}\ = \frac 2 3$

2. Originally Posted by Creebe
How do I prove: $\sum_{n=1}^{\infty} \frac {(-1)^n} {2^n}\ = \frac 2 3$
Prove it? Show by induction that $\sum_{k=1}^{n}\left(\tfrac{-1}{2}\right)=\frac{\left(\tfrac{-1}{2}\right)^{n+1}-1}{\tfrac{-1}{2}-1}$ and then prove that $\lim\frac{\left(\tfrac{-1}{2}\right)^{n+1}-1}{\tfrac{-1}{2}-1}=\frac{-1}{\tfrac{-1}{2}-1}$

3. thanks! but could you explain this part?

$\sum_{k=1}^{n}\left(\tfrac{-1}{2}\right)=\frac{\left(\tfrac{-1}{2}\right)^{n+1}-1}{\tfrac{-1}{2}-1}$

4. Originally Posted by Creebe
How do I prove: $\sum_{n=1}^{\infty} \frac {(-1)^n} {2^n}\ = \frac 2 3$
Have you ever seen the geometric series? This is just a particular case with $x=-\frac12.$

5. Here is what Krizalild is talking about:

Let $S= \sum_{i=0}^n ar^{i}= a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^n$

Write that as $S= a+ r(a+ ar+ \cdot\cdot\cdot+ ar^{n-1})$
Where I have separated the first "a" to the left side of the equation and factored an "r" out of the rest. The expression still inside the parentheses is almost what we started with- it is missing only the last term.

So put that back in by "adding and subtracting the same thing":
$S= a+ r(a+ ar+ \cdot\cdot\cdot+ ar^{n-1}+ ar^n- ar^n)$
$S= a+ r(a+ ar+ \cdot\cdot\cdot+ ar^{n-1}+ ar^n)- ar^{n+1}$

Now, I have moved that last " $-ar^{n+1}$ outside the parenthese, multiplying it by the r that multiplies the entire parentheses, of course.

But what is inside the parentheses is exactly the sum we started with- it is S.
$S= a+ rS- ar^{n+1}$
Now solve for S. Subtract rs from both sides to get $S- rS= (1-r)S= a- ar^{n+1}$ and then divide both sies by 1- r: $S= \frac{a(1- r^{n+1})}{1- r}$.
(In order to divide by 1- r, r must not be 1, of course. But if r= 1, the sum is just $\sum_{i=0}^n a= a(n+1)$.)

That is the formula for the "finite geometric series". To get the formula for the infinite geometric series, take the limit as n goes to infinity. If |r|> 1, $r^{n+1}$ will get larger and larger and there is no limit. If r= 1, this is just a+ a+ a+ ...+ a= (n+1)a which has no limit. If r= -1, this is just a- a+ a- a+...+ a which is either 0 or a depending on whether n is even or odd and so there is no limit. But if |r|< 1, $r^{n+1}$ goes to 0 so we have
$\sum_{n=0}^\infty ar^n= \frac{a}{1- r}$
as long as |r|< 1.

In this particular problem, $r= -\frac{1}{2}$ so $|r|= \frac{1}{2}< 1$ and the sum is
$\frac{1}{1-(-\frac{1}{2})}= \frac{1}{\frac{3}{2}}= \frac{2}{3}$