Question

How do you find the radius of convergence `Sigma x^n/3^n` from `n=[0,oo)`?

Answer 1

The series has radius on convergence `R=3`, that is:

`sum_(n=0)^oo x^n/3^n = 3/(3-x)` for `abs(x) < 3`

Explanation:

We can write the series as:

`sum_(n=0)^oo x^n/3^n = sum_(n=0)^oo (x/3)^n = sum_(n=0)^oo t^n`

where `t=x/3`.

Now this is the geometrical series of ratio `t` and we should know that it has radius of convergence `R=1`, so that it converges absolutely for:

`abs (t) < 1 => abs (x/3) < 1 => abs (x) <3`

and the sum is:

`sum_(n=0)^oo x^n/3^n =1/(1-x/3)= 3/(3-x)`

Answer 2

The radius of convergence is `=3`

Explanation:

We use the series ratio test

`∣a_(n+1)/a_n∣=(∣(x^(n+1)/(3^(n+1)))/(x^n/3^n)∣)`

`lim_(n->+oo)∣x^(n+1)/(x^n)*3^n/3^(n+1)∣`

`=∣x∣*1/3`

For the series to converge, we need that

`∣x∣*1/3<1`

Therefore,

`∣x∣<3`

The series converge for `-3 < x <3`

For, `x=3`, `=>`, `sum_0^(+oo)3^n/3^n`

`=sum_0^(+oo)1`, the series diverge

For, `x=-3`, `=>`, `sum_0^(+oo)(-3)^n/3^n`

`=sum_0^(+oo)(-1)^n`, the series diverge