Evaluate sigma question data

[Felicia123]

s = sigma ( i / 4^i ) for i = 0 to infinite
s = 1/4 + 2/4^2 + 3/4^3 + ... (1)
4s = 1 + 2/4 + 3/4^2 + ... (2)
Subtract the second from the third,
3s = 1 + 1/4 + 1/4^2 + ... = 1/(1 - 1/4) = 4/3
s = 4/9 <==Answer

can i ask
isnt (2)-(1)?

but how to get
3s= 1 + 1/4 + 1/4^2 from there ?
i can't get it for 3s there only

[MiiNiPaa]

1 + 2/4 + 3/4^2 + 4/4^3 + 5/4^4 + ...
−
    1/4 + 2/4^2 + 3/4^3 + 4/4^4 + ...
=
1 + 1/4 + 1/4^2 + 1/4^3 + 1/4^4 + ...

[Felicia123]

thanks for your help !
i get it finally

by the way.
then how they get the

1/(1 - 1/4) = 4/3
already?
the answer until infinity..

[MiiNiPaa]

http://en.wikipedia.org/wiki/Geometric_series#Sum
Formula: a/(1 − r)
in your case:
a = 1;
r = 1/4;