Answer:
We can see that:
a₀ = 4
a₁ = 4/3
a₂ = 4/9.
We easily can see that the n-th therm of this will be:
aₙ = 4/3^n.
Now, the denominator increases as n increases, then for a really large n, aₙ will tend asymptotically to zero. This means that this summation converges.
We can write this summation as:
4*∑(1/3)^n.
The sum for the first N-th therms is:
We know that for a summation:
1 + r + r^2 + r^3 + ... + r^N = (1 - r^(N + 1))/(1 - r)
4 + 4*(1/3) + 4*(1/3)^2 + ... + 4*(1/3)^N = 4*( 1 - (1/3)^(N + 1))/(1 - (1/3))
Now, for the complete sum we have that:
n = {0, 1, 2, ...}
We know that for a summation:
∑a*r^n
with n = {0, 1, ...}
the sum is = a/(1 + r).
in this case we have:
a = 4, r = 1/3.
Then the sum is:
S = 4/(1 - 1/3) = 4/(2/3) = 3*2 = 6
