What's the sum of the series from zero to infinity of ((-1)^(n+1))((4)^(3-n))?

Question

The actual problem is stated as "Find the sum of the convergent series: 64 - 16 +4 -1 +..." and I derived the series notation stated above in the question. I think it's right, please correct me if I'm wrong. But if correct, how do I find the sum? Don't I need to take the limit of some sort of notation of the sum? The series appears to be geometric and if it was I know that the sum of a geometric series is equal to (a/1-r) however, it looks like the "ratio" in this case is 4 but that would mean that it'd be divergent however the problem states it's convergent. And also I have been given the answer to be 256/5 but i do not know how to derive it. Please help!

Answer 1

The nth term of the series can also be written as



((-1)^(n-1))((4)^(4 - n)) [for n =1 we should get 64 and not 16 ]



((-1)^(n-1))((4)^(3 - n + 1)) = ((-1)^(n - 1))((4^3)(4^(1-n)))



= 64 {(-1/4) ^ (n-1)} [in the form a r^(n-1)]



Thus it is a geometric series with a=64 and r = -1/4



sum = 64 / (1+1/4) = 256/5

Answer 2

It's the sum of a geometric sequence with

initial term a[0] = 64 and

common ratio r = -1/4.



Since you want to start counting at 0 {good for you!}, then the

nth term will be a[n] = 64(-1/4)^n



The sum will be a[0] / (1 - r) = 64/(5/4) = 64 * (4/5) = 256/5 = 51.2

Answer 3

The initial term is a = 64, and the ratio is r = -1/4.

So, the sum converges to a/(1 - r) = 64/(1 - (-1/4)) = 256/5.



I hope this helps!