Please help me with this series question.. So evaluate the sum from n=0 to infinity of n/4^n....I cant figure out the common ratio.?

This is not a geometric series, but nyc_kid's solution expresses it as the derivative of one.

The first 20 terms are:

0.25000000000000

0.12500000000000

0.04687500000000

0.01562500000000

0.00488281250000

0.00146484375000

0.00042724609375

0.00012207031250

0.00003433227539

0.00000953674316

0.00000262260437

0.00000071525574

0.00000019371510

0.00000005215406

0.00000001396984

0.00000000372529

0.00000000098953

0.00000000026193

0.00000000006912

0.00000000001819

And that sum is

0.44444444443798 suggesting strongly (but not proving) that 4/9 is the answer.

Consider the function f(x) = 1 + (x/4) + (x/4)^2 + (x/4)^3 +....... = 4/(4 -x) for all x in [0, 4)

hen, f'(1) = 1/4 + 2/4^2 + 3/4^3 + ....... = [ 4/(4 -x) ]' at x=1 .

But [ 4/(4 -x) ]' = 4/(4-x)^2 which equals 4/9 at x=1.

So, the answer is 4/9.

Now what common ratio are you talking about????