262S_{n} = \sum_{1}^{n}{\frac{1}{2^{r}}}
S_{n} = \frac{1}{2} + \frac{1}{2^{2}} + .... + \frac{1}{2^{n}}
S_{n} = \frac{1}{2}*\frac{1- (1/2)^{n}}{1-(1/2)}
S_{n} = 1- (1/2)^{n}
0<S_{\propto } <1
\left[ S_{\propto }\right] =0
11ya i had a gross mistake but are u sure the limit isn't 1?
1357another way
S=(1/2)+(1/22).....∞
so it is infinite GP.
S=a/(1-r)=(1/2)/(1-1/2)=1
2305It's not 0 it's 1.In the inequality you had given , you didn't consider n=∞.
As n→∞ (1/2)n→0
Hence 0<Sn≤1
And S∞=1
