Find the sum of the series

Summation of infinity, n = 0

2n/(8^((2n) +1))


The next question is:

(4^n)/(5^n) where n = 5

1) Start with the geometric series
1/(1 - x) = Σ(n = 0 to ∞) x^n.

Differentiate both sides:
1/(1 - x)^2 = Σ(n = 1 to ∞) nx^(n-1).

Multiply both sides by x:
x/(1 - x)^2 = Σ(n = 1 to ∞) nx^n.
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So, Σ(n = 0 to ∞) 2n / 8^(2n+1)
= (2/8) * Σ(n = 0 to ∞) n/8^(2n)
= (1/4) * Σ(n = 1 to ∞) n (1/64)^n
= (1/4) * (1/64)/(1 - 1/64)^2
= 16 / 3969.
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2) Σ(n = 5 to ∞) (4/5)^n
= (4/5)^5 / (1 - 4/5)
= 1024/625.

I hope this helps!