We should calculate the radius of convergence of the series
"\\sum\\limits_{n=0}^{\\infty} \\dfrac{nx^n}{3^n} = \\sum\\limits_{n=0}^{\\infty} n \\left(\\dfrac{x}{3}\\right)^n."
We may use the D'Alembert's rule and calculate the radius of convergence as
"R = \\lim\\limits_{n\\to\\infty} \\left| \\dfrac{n \\left(\\dfrac{1}{3}\\right)^n}{(n+1) \\left(\\dfrac{1}{3}\\right)^{n+1}} \\right| = 3."
If "x=3," then "\\sum\\limits_{n=0}^{\\infty} n\\left(\\dfrac{x}{3}\\right)^n = \\sum\\limits_{n=0}^{\\infty} n = +\\infty."
If "x=-3," then "\\sum\\limits_{n=0}^{\\infty} n\\left(\\dfrac{x}{3}\\right)^n = \\sum\\limits_{n=0}^{\\infty} (-1)^nn," this series is not convergent, therefore the interval of convergence is (-3, 3).
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