1. ## clearification on series

why is

$\displaystyle \sum^{\infty}_{x=1} e^{tx} \left( \frac{1}{3} \right)^{x} = \sum^{\infty}_{x=0} \left( \frac{e^t}{3} \right)^{x} -1$

I get that X which was originally equal to is now equal to 0 but wouldn;t this imply:

$\displaystyle \sum^{\infty}_{x=1} e^{tx} \left( \frac{1}{3} \right)^{x} = \sum^{\infty}_{x=0} e^{t(x+1)}\left( \frac{1}{3} \right)^{x+1}$

how do you get rid of the $\displaystyle +1$ to get -1?

2. $\displaystyle \sum\limits_{x\, = \,1}^\infty {e^{tx} \left( {\frac{1} {3}} \right)^x } = \sum\limits_{x\, = \,1}^\infty {\left( {\frac{{e^t }} {3}} \right)^x } = \frac{{e^t }} {3}\sum\limits_{x\, = \,1}^\infty {\left( {\frac{{e^t }} {3}} \right)^{x - 1} } = \frac{{e^t }} {3}\sum\limits_{x\, = \,0}^\infty {\left( {\frac{{e^t }} {3}} \right)^x } .$

3. I think you just confusing your self here.

This i just basic rules of indices.

$\displaystyle \sum^{\infty}_{x=1} e^{tx} \left( \frac{1}{3} \right)^{x} = \sum^{\infty}_{x=1} \left( \frac{e^t}{3} \right)^{x}$

your thing goes down to the zeroth term, because you add an extra term you take it away to keep equality.

$\displaystyle \sum^{\infty}_{x=1} \left( \frac{e^t}{3} \right)^{x} = \sum^{\infty}_{x=0} \left( \frac{e^t}{3} \right)^{x} - \left( \frac{e^t}{3} \right)^{0}$

$\displaystyle \Rightarrow \sum^{\infty}_{x=0} \left( \frac{e^t}{3} \right)^{x} - 1$

4. would there be any point on going from 1 to 0?

5. ## Yes

Originally Posted by lllll
would there be any point on going from 1 to 0?
there are many reasons why you would want to change the indices...one to easier find the sum of series...also so you may combine through addition, subtraction, etc. series with different indices

6. Originally Posted by lllll
would there be any point on going from 1 to 0?
No. Unless $\displaystyle x \geq 0$ was inherent in the problem in which case it's essential to start from x = 0.