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I need to find the radius of convergence of $\Sigma n^3z^n$

I want to use the ratio test because it would be simpler than the root test. If $C_n=n^3$ then $| \dfrac {C_{n+1}}{C_n}| > 1$ because $(n+1)^3 > n^3$ for $\forall n>0$.

The problem I face is that the solution manual claims that $| \dfrac {C_{n+1}}{C_n}| = 1$ which means the radius of convergence would be $\dfrac {1}{1}=1$.

Will someone please explain to me why this is?

Is it because both the numerator and denominator are both approaching infinity? I assumed this was wrong because $\dfrac{\infty}{\infty}$ is undefined even if intuitively it seems like it would equal 1. Also, the numerator is approaching infinity significantly faster.

blubberbrot
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The quotient $|C_{n + 1} / C_n|$ is never equal to $1$, as you've observed. The important fact, however, is that the limit

$$\lim_{n \to \infty} \left|\frac{C_{n + 1}}{C_n}\right| = 1$$

This follows from the fact that

$$\frac{(n + 1)^3}{n^3} = 1 + \frac 3 n + \frac 3 {n^2} + \frac 1 {n^3} \to 1 + 0 + 0 + 0 = 1$$

as $n \to \infty$.