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

2 Answers
Andrea S. · Steve M
Jan 11, 2017

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)

Narad T.
Jan 11, 2017

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

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