find a formula for the nth partial sum of each series 4+(4/3)+(4/9) and use it to find the series’ sum of the series converges.

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

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step-by-step explanation: