limits - Find $lim_{n to infty} frac{(-1)^n n^{frac{3}{2}}}{5e^n}$

Question: Find $\lim_{n \to \infty} \frac{(-1)^n n^{\frac{3}{2}}}{5e^n}$

My attempt: Since $(-1)^n$ can be written as $\cos(\pi n)$ we can use the squeeze theorem as such: $-1 \leq \cos(\pi n) \leq 1 \implies \frac{-n^{\frac{3}{2}}}{5e^n} \leq \frac{\cos(\pi n)\cdot n^{\frac{3}{2}}}{5e^n}\leq \frac{n^{\frac{3}{2}}}{5e^n}$ and so we can take the limit as $n$ approaches zero and because exponentials grow faster than polynomials we get: $0\leq \frac{\cos(\pi n)\cdot n^{\frac{3}{2}}}{5e^n}\leq 0$ so $\lim_{n \to \infty} \frac{(-1)^n n^{\frac{3}{2}}}{5e^n} =0$.

Is this correct? If so what other ways could I have solved this limit?

Answer

alternative method:-$|a_n|=\frac{n^{\frac{3}{2}}}{5e^n}$. Consider the series $\sum_{n=0}^{\infty}|a_n|$. Use cauchy's nth root test, we get $\sum_{n=0}^{\infty}|a_n|$ converges. so $\lim_{n\to\infty}|a_n|=0.$ Hence, $\lim_{n\to\infty}a_n=0$.