What am I doing wrong with this proof?

• August 16th 2011, 10:35 AM
terrorsquid
What am I doing wrong with this proof?
So, I'm trying to use induction to prove:

$\sum_{k=0}^n\left(\frac{-1}{3}\right)^k = \frac{3}{4} + \frac{1}{4}(-1)^n3^{-n},~~$ $~~for~all~n \in \mathbb{N}$

Base case is true.

Assume:

$\left(\frac{-1}{3}\right)^k = \frac{3}{4} + \frac{1}{4}(-1)^n3^{-n}$

Test:

$\left(\frac{-1}{3}\right)^{k+1} = \frac{3}{4} + \frac{1}{4}(-1)^{(n+1)}3^{-(n+1)}$

$\left(\frac{-1}{3}\right)^k\left(\frac{-1}{3}\right) = \frac{3}{4} + \frac{1}{4}(-1)^{n}(-1)3^{-n}3^{-1}$

$\left(\frac{3}{4} + \frac{1}{4}(-1)^n3^{-n}\right)\left(\frac{-1}{3}\right) = \frac{3}{4} -\frac{1}{12}(-1)^{n}3^{-n}$

This next step is what seems to cause the problems. If the 3/4 on the LHS was not in brackets, everything would be fine and become = when I multiply the other term by -1/3; however, by multiplying the 3/4 by -1/3 it makes them not equal:

$-\frac{1}{4} - \frac{1}{12}(-1)^n3^{-n} = \frac{3}{4} -\frac{1}{12}(-1)^{n}3^{-n}$

What am I doing wrong?
• August 16th 2011, 03:45 PM
topspin1617
Re: What am I doing wrong with this proof?
What happened to all of the summation signs??

You're trying to prove something about the entire summation $\sum_{k=0}^n\left$$\frac{-1}{3}\right$$^k$, not the individual term $\left$$\frac{-1}{3}\right$$^k$.
• August 16th 2011, 04:08 PM
Plato
Re: What am I doing wrong with this proof?
Quote:

Originally Posted by terrorsquid
So, I'm trying to use induction to prove:
$\sum_{k=0}^n\left(\frac{-1}{3}\right)^k = \frac{3}{4} + \frac{1}{4}(-1)^n3^{-n},~~$ $~~for~all~n \in \mathbb{N}$

Without induction.
Let $S_n = \sum\limits_{k = 0}^n {r^k }$

\begin{align*} S_n&= 1+r+r^2+\cdots+r^n \\ rS_n &=r+r^2+r^3\cdots+r^{n+1}\\(1-r)S_n &=1-r^{n+1}\\ S_n &= \frac{1-r^{n+1}}{1-r} \end{align*}
• August 16th 2011, 08:15 PM
terrorsquid
Re: What am I doing wrong with this proof?
Quote:

Originally Posted by topspin1617
What happened to all of the summation signs??

It was a magic sum sign :D

haha, stayed up too late last night I think.

Thanks.