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

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Answer 1 · 2017-01-11T15:06:04

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#

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 · 2017-01-11T15:21:30

The radius of convergence is #=3#

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