# Value of S?

• Nov 11th 2007, 05:53 AM
Simplicity
Value of S?
Could someone help with this, A hint is provided but I cannot seems to find a solution.
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$\displaystyle S = 1 + 0.5 + 0.25 + ... + (\frac{1}{2})^n$

What is the Value of S? [HINT = Multiply S by 2, then Solve]
• Nov 11th 2007, 06:01 AM
kalagota
Quote:

Originally Posted by Air
Could someone help with this, A hint is provided but I cannot seems to find a solution.
__________________________________________________ _______________________

$\displaystyle S = 1 + 0.5 + 0.25 + ... + (\frac{1}{2})^n$

What is the Value of S? [HINT = Multiply S by 2, then Solve]

you have actually this
$\displaystyle S = 1 + \frac{1}{2} + \left( {\frac{1}{2}} \right)^2 + ... + \left( \frac{1}{2} \right)^n$ --------(a)

if we multiply 2, we have

$\displaystyle 2S = 2 + 1 + \frac{1}{2} + \left( {\frac{1}{2}} \right)^2 + ... + \left( \frac{1}{2} \right)^{n-1}$ ----------(b)

subtracting (b) - (a), we have

$\displaystyle S = 2 - \left( \frac{1}{2} \right)^n$, so if we take n to be a large number, then $\displaystyle S \approx 2$
• Nov 11th 2007, 06:05 AM
Simplicity
Quote:

Originally Posted by kalagota
$\displaystyle S = 2 - \left( \frac{1}{2} \right)^n$, so if we take n to be a large number, then $\displaystyle S \approx 2$

What do you mean by n is larger number? :o Can you explain how the approximation of 2 is obtained?
• Nov 11th 2007, 07:48 AM
earboth
Quote:

Originally Posted by Air
What do you mean by n is larger number? :o Can you explain how the approximation of 2 is obtained?

Hello,

you are supposed to know that

$\displaystyle \lim_{n\to \infty}\left(\frac12\right)^n = 0$

With your problem you have:

$\displaystyle \lim_{n\to \infty}\left(2- \left(\frac12\right)^n \right)= 2-0 = 2$