Casework using ratio test.

Question

We have to find the limit of the sequence $ (a_n) = \frac{y^n}{n^m}$ as $n \to \infty$, for all values of $y > 0$ and $m = 1,2,...$

Now using the ratio test: $\frac{a_{n+1}}{a_n} = y(\frac{n}{n+1})^m$ where $\frac{n}{n+1} \to 1$ as $n \to \infty$. Therefore $(\frac{a_{n+1}}{a_n}) \to y$ as $n \to \infty$.

The limit can only exist if $0<y<1$ (in which case it is zero), but I am unsure what I can say about the case where $y \ge 1$ and how to tackle this case.

Answer 1

Recall that by ratio test

$$\frac{a_{n+1}}{a_n} \to L>1 \implies a_n \to \infty$$

indeed by the definition of limit we have that eventually $\forall n\ge n_0$

$$\frac{a_{n+1}}{a_n}>\frac {1+L}2 \implies a_{n_0+k}=\left(\frac {1+L}2\right)^ka_{n_0} \implies a_n=\left(\frac {1+L}2\right)^{n-n_0}a_{n_0} \to \infty$$

Answer 2

Hint : $$\ln\bigg(\frac{y^n}{n^m}\bigg) = n\ln(y) - m \ln(n)$$