Find the radius and interval of convergence for the power series $$\displaystyle{\sum_{n=0}^{\infty}}\left( \frac{n!}{2 \times 5 \times 8 \times \ldots\times(3n+2)} \right)^3x^{5n}$$


Hint. Let $a_n=\left(\frac{n!}{2 \times 5 \times 8 \times \cdots\times(3n+2)}\right)^3$ then $$\lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to +\infty}\left(\frac{n+1}{3n+5}\right)^3=\frac{1}{27}.$$ Hence, by the Ratio test, the radius of convergence of the power series $$z\to\sum_{n=0}^{\infty}a_nz^n$$ is $R=27$.

Then, what is the radius of convergence of the power series $$x\to\sum_{n=0}^{\infty}a_nx^{5n}\quad ?$$

As regards the product $2 \times 5 \times 8 \times \cdots \times (3n+2)$ see also $f_2(n)$ in closed-form expressions for product of 3n+k where k = 1 or 2 You need the Stirling's approximation to determine the exact interval of convergence.

  • $\begingroup$ yes, thank you !I have one question tho.The only thing that bothers me is what do I do with the 2x5x8....? $\endgroup$ – Alexisdr Jun 17 '19 at 14:35
  • $\begingroup$ Have you tried to compute the ratio $\frac{a_{n+1}}{a_n}$? $\endgroup$ – Robert Z Jun 17 '19 at 14:38

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