topsquark
Senior Member
- Joined
- Aug 27, 2012
- Messages
- 2,370
They never do!But then I am still confused about how the segments/terms reach the other side. A limit just gets arbitrarily close doesn't it? In other words, each n only gets the sum close. How does the gap get filled?
The Nth partial sum is
n=1∑N2n1=21⎝⎜⎜⎜⎜⎛1−211−(21)N⎠⎟⎟⎟⎟⎞=1−(21)N
Now, we want this in the limit as N goes to infinity. Well, nothing happens to the 1. What happens to the (1/2)N? Notice that 0 < 1/2 < 1. Every time we put a higher exponent on this, the smaller it gets. If we do this an arbitrarily large number of times, we get arbitrarily closer to 0. So
N→∞lim(21)N=0
Thus
n=1∑∞2n1=1−0=1
If you want to do the limit for real, look up epsilon-delta proofs. But the point is that we never took a value for N to find this. We simply noted that the bigger N is, the closer (1/2)N gets to 0. This is not the sort of process where you can imagine that you are summing an very large number of terms. We sum them for an arbitrary N and then we take a limit, which does the job for us.
-Dan