Question

Help with a series
\[\sum_{i=0}^{\infty}\left(\frac{1}{2}\right)^i\]

How to find its value.

Answer 1

1,1/2,1/4,....
thats a geometric sequence with infinite terms and common ratio =1/2
do u realize that ?

Answer 2

the formula to calculate sum of infinite geometric sequence is a/(1-r)
where r=1/2 = common ratio and a=1st term =1

Answer 3

yes, I realize it's a geometric sequence. I see that is the formula now. So, for the sequence above, the answer will be 2.

I suppose it is quite a long explanation as to why the formula is 1/(1-r)?

Answer 4

I mean, \[\frac{a}{1-r}\]

Answer 5

not really

Answer 6

yes, its 2.
it id derived by putting n-> infinity in the general sum formula for n terms.

Answer 7

it is to do it "rigorously" but to see it is simple enough
forget about the \(a\) you can always factor it out and consider only
\[1+r+r^2+r^3+...+r^n\]
multiply by \(1-r\) and see that you get \(1-r^{n+1}\) meaning
\[1+r+r^2+...+r^n=\frac{1-r^{n+1}}{1-r}\] as \(n\to \infty\) we get \(r^{n+1}\to 0\) provided of course \(|r|<1\)

Answer 8

and @satellite73 proved it nicely :)

Answer 9

second method
call the sum \(S=1+r+r^2+r^3+...\) then \(rS=r+r^2+r^3+...\) so
\[S-rS=1\] and thus \((1-r)S=1\)making \[S=\frac{1}{1-r}\]

Answer 10

all this stuff begs the question about what it means to add an infinite number of numbers
to do it for real you need to be more precise and talk about limits etc, but the algebra or visualization is not too arcane
@hartnn thanks!

Answer 11

Yes, thank you both very much, @hartnn for getting me started, and @satellite73 for exhaustively explaining enough "proof" for my needs. Perhaps in a few years I'll be ready for more rigorous proofs. The second method was especially easy to visualize for me.

Answer 12

yw