How do you find the radius of convergence #Sigma x^(3n)/5^n# from #n=[1,oo)#?

#sum_(n=1)^oo x^(3n)/5^n = sum_(n=1)^oo (x^3/5)^n#

So, this is a geometric series of ratio: #x^3/5#, which is convergent for:

#-1 <= x^3/5 < 1#

#-5 <= x^3 < 5#

#-root(3)5 <= x < root(3)5#

To find the radius of convergence for the series \( \sum_{n=1}^{\infty} \frac{x^{3n}}{5^n} \), you can use the ratio test. The ratio test states that if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \), then the series converges if \( L < 1 \) and diverges if \( L > 1 \). For the given series, \( a_n = \frac{x^{3n}}{5^n} \). Thus, \( a_{n+1} = \frac{x^{3(n+1)}}{5^{n+1}} = \frac{x^{3n+3}}{5^{n+1}} \). Taking the ratio of consecutive terms: \[ \frac{a_{n+1}}{a_n} = \frac{\frac{x^{3n+3}}{5^{n+1}}}{\frac{x^{3n}}{5^n}} = \frac{x^{3n+3} \cdot 5^n}{5^{n+1} \cdot x^{3n}} = \frac{x^3}{5} \] As \( n \) approaches infinity, this ratio becomes \( \frac{x^3}{5} \). Now, taking the limit as \( n \) approaches infinity: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{x^3}{5} \right| \] For the series to converge, this limit must be less than 1. Hence, solve: \[ \left| \frac{x^3}{5} \right| < 1 \] This simplifies to: \[ |x^3| < 5 \] Thus, \( -5 < x^3 < 5 \). Since the series will converge if \( |x^3| < 5 \), this means that \( x \) must lie within the interval \( (-5^{1/3}, 5^{1/3}) \). Hence, the radius of convergence is \( 5^{1/3} \).