How do you find the interval of convergence #Sigma n3^nx^n# from #n=[0,oo)#?

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Answer 1 · 2017-02-22T23:06:23

The series

#sum_(n=0)^oo n3^nx^n#

is convergent for #x in (-1/3,1/3)#

Given the series:

#sum_(n=0)^oo n3^nx^n#

We can apply the ratio test to determine the interval of values of #x# for which the series is convergent.

We then evaluate:

#abs (a_(n+1)/a_n) = abs ( ( (n+1)3^(n+1)x^(n+1))/(n3^nx^n)) = 3((n+1)/n)absx#

So we have that:

#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) 3((n+1)/n)absx = 3 absx#

The series is then absolutely convergent for #absx < 1/3# and divergent for #abs x > 1/3#

In the cases where #abs x = 1/3 # the ration test is indecisive and we have to analyze in detail:

(i) #x = 1/3#

#sum_(n=0)^oo n3^nx^n = sum_(n=0)^oo n3^n(1/3)^n = sum_(n=0)^oo n = oo#

(iI) #x = -1/3#

#sum_(n=0)^oo n3^nx^n = sum_(n=0)^oo n3^n(-1/3)^n = sum_(n=0)^oo (-1)^n n #

Now if we can consider the partial sums of even order, we have:

#s_(2N) = sum_(n=0)^(2N) (-1)^n n = sum_(n=0)^N 2n-(2n-1) = N#

while for the partial sums of odd order:

#s_(2N+1) = s_(2N) -(2N+1) = N-2N-1 = -N-1#

so the series is irregular.

In conclusion the series is convergent for #x in (-1/3,1/3)#