We use the ratio test to determine what the interval of convergence is.
The ratio test states that a series #sum a_n# converges if #lim_(n->∞) abs(a_(n+1)/a_n) < 1#
We can substitute our given series into this test, #lim_(n->∞) abs((x^(n+1)/((n+1)(3^(n+1))))/((x^n)/(n3^n))) = abs((x^(n+1)*n*3^n)/((n+1)(3^(n+1)*x^n)#
Simplifying, we get: #lim_(n->∞)abs((x^(1)*n)/((n+1)(3^(1)))) = lim_(n->∞)abs((x*n)/((3n+3)))#
We can pull out the #n# from the expression like so:
#x*lim_(n->∞)abs((n)/((3n+3))#
We can divide all terms in the limit by #n# to make evaluating it easier:
#x*lim_(n->∞)abs((n/n)/((3n/n+3/n))) = x*lim_(n->∞)abs((1)/((3+3/n))#
Evaluating the limit, we get #x*lim_(n->∞)abs((1)/((3+3/n))) = 1/3#
We can now distribute #x# back into the evaluated limit, thus we get that #abs(x/3) < 1# if the series is to converge.
This means that either #x/3 < 1 or x/3 > -1#, that is #x < 3 or x > -3#
We must now check if #x# will actually converge at those end points too, substituting 3 as x into the original series, we get: #sum 3^n/(n3^n) = sum 1/n(3/3)^n = sum 1/n 1^n = sum 1/n# this series is not convergent by the p-series test. So #x=3# is non convergent.
Now we test #x=-3#, doing this we get: #sum (-3)^n/(n3^n) = sum ((-1)^n(3)^n)/(n3^n) = sum(-1)^n/n#, now this series is actually convergent by the alternating series test.
Thus our interval of convergence is #[-3,3)#
