$\displaystyle \sum\limits_{n = 0}^{n - 1} \left({\frac {1}{2}}\right) ^ n$

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- Feb 27th 2008, 10:05 AMEquinoXhelp me to simplify this
$\displaystyle \sum\limits_{n = 0}^{n - 1} \left({\frac {1}{2}}\right) ^ n$

- Feb 27th 2008, 11:14 AMwingless
I assume that $\displaystyle n-1$ on sigma is a constant and not related to the summation.

$\displaystyle \sum _{k=0}^a r^k = \frac{r^{1+a}-1}{r-1}$

So,

$\displaystyle \sum_{n = 0}^{n - 1} \left({\frac {1}{2}}\right)^n = \frac{\left(\frac{1}{2}\right)^{1+(n-1)}-1}{\frac{1}{2}-1} = 2 - 2^{1-n}$ - Feb 28th 2008, 03:02 AMgen2danwel wel wel
well i think the answer to that question is 1 though im not sure

- Feb 28th 2008, 03:44 AMmr fantastic
If it was the infinite series $\displaystyle \sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^n$ then the answer would indeed be 1.

But it's not so it's not.

(Note: The very first term of the given series is $\displaystyle \left( \frac{1}{2} \right)^0 = 1$ so clearly it can't sum to 1 .......)