Question

How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given `1+1/3+1/9+...+(1/3)^n+...`?

Answer 1

`S_n =sum_(k=0)^n (1/3)^k = 1/2 (3^n-1)/3^(n-1)`

`S= sum_(k=0)^oo (1/3)^k = 3/2`

Explanation:

This is a geometric series of ratio `q=1/3`:

`S= sum_(k=0)^oo (1/3)^k = sum_(k=0)^oo q^k`

Consider the `n`-th partial sum of the series:

`S_n = sum_(k=0)^(n-1) q^k = 1+q+q^2+...+q^(n-1)`

and note that:

`(1+q+q^2+...+q^(n-1))(1-q) = 1-cancelq+cancelq -cancel(q^2)+cancel(q^2) +...-q^n =1-q^n`

so that:

`S_n = (1-q^n)/(1-q) = (1-1/3^n)/(1-1/3) = 3/2(3^n-1)/3^n = 1/2 (3^n-1)/3^(n-1)`

As `lim_(n->oo) q^n = lim_(n->oo) (1/3)^n = 0` we have

`lim_(n->oo) S_n = (1-q^n)/(1-q) = 1/(1-q) = 1/(1-1/3) = 3/2`

And in general we can see that a geometric series converges when:

`lim_(n->oo) q^n = 0`

that is for `absq < 1`