Question

Evaluate the sum:

[(infinity) sigma i=0] (i/(4^i))

the answer is 4/9, but what's the step?

Thank you.

Answer 1

you want the sum of
0/4^0 + 1/4^1 + 2/4^2 + 3/4^3 + ...
= 0 + 1/4 + 2/16 + 3/64 ..

sum(0) = 0
sum(1) = 0+1/4 = 1/4
sum(2) = 1/4 + 2/16 = 6/16
sum(3) = 3/8 + 3/64 = 27/64
sum(4) = 27/64 + 4/256 = 112/256
..
looks like the denominators follow the pattern
sum(n) = ??/4^n

so how about the top?
0, 1, 6, 27, 112, ....

I can't seem to find a pattern, unless I am missing something obvious.

so far we have sum(4) = 112/256 = .4375
it looks like we do approach 4/9 which is .4444...

Answer 2

The top seems to be an arithmetic sequence and the bottom a geometric sequence, but I can not get that ratio to converge as i --->oo