Question

What is the sum of `2/3+8/9+26/27+...` to `n` terms ?

Answer 1

`sum_(k=1)^n a_k = n - 1/2 + 1/2(1/3)^n`

Explanation:

The general term of a geometric series with initial term `a` and common ratio `r` can be written:

`a_k = a*r^(k-1)`

The sum to `n` terms is given (see footnote) by the formula:

`sum_(k=1)^n = (a(1-r^n))/(1-r)`

Note that:

`2/3 = 1 - 1/3`

`8/9 = 1 - 1/9`

`26/27 = 1 - 1/27`

So it seems that the general term of the given series is:

`a_k = 1 - 1/3^k`

We have:

`sum_(k=1)^n 1/3^k = (1/3(1-(1/3)^n))/(1-1/3)`

`color(white)(sum_(k=1)^n 1/3^k) = (1-(1/3)^n)/2`

`color(white)(sum_(k=1)^n 1/3^k) = 1/2-1/2(1/3)^n`

So:

`sum_(k=1)^n a_k = sum_(k=1)^n (1-1/3^k)`

`color(white)(sum_(k=1)^n a_k) = sum_(k=1)^n 1 - sum_(k=1)^n 1/3^k`

`color(white)(sum_(k=1)^n a_k) = n - (1/2-1/2(1/3)^n)`

`color(white)(sum_(k=1)^n a_k) = n - 1/2 + 1/2(1/3)^n`

`color(white)()`
Footnote

Given:

`a_k = ar^(k-1)`

We find:

`(1-r)sum_(k=1)^n a_k = (1-r)sum_(k=1)^n ar^(k-1)`

`color(white)((1-r)sum_(k=1)^n a_k) = sum_(k=1)^n ar^(k-1) - rsum_(k=1)^n ar^(k-1)`

`color(white)((1-r)sum_(k=1)^n a_k) = sum_(k=1)^n ar^(k-1) - sum_(k=2)^(n+1) ar^(k-1)`

`color(white)((1-r)sum_(k=1)^n a_k) = a+color(red)(cancel(color(black)(sum_(k=2)^n ar^(k-1)))) - color(red)(cancel(color(black)(sum_(k=2)^n ar^(k-1)))) - ar^n`

`color(white)((1-r)sum_(k=1)^n a_k) = a(1-r^n)`

So dividing both ends by `(1-r)` we get:

`sum_(k=1)^n a_k = (a(1-r^n))/(1-r)`