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

Question

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?

The reasoning is correct, but I wouldn't write $0 \leq \textrm{cosine term} \leq 0$, this is not alright. — max_zorn, Jan 01 '18 at 06:40
You don't need the cosine to say that $-1\le (-1)^n \le 1$. Also, a powerful way to prove that $u_n\to 0$ is to prove that $|u_n|\to 0$. You could also consider $\frac{|u_{n+1}|}{|u_n|}$ — Gribouillis, Jan 01 '18 at 07:00
Possible duplicate of How to prove that exponential grows faster than polynomial? — Guy Fsone, Jan 01 '18 at 08:55
https://math.stackexchange.com/questions/266640/prove-lim-n-rightarrow-infty-fracbnnk-infty?rq=1 — Guy Fsone, Jan 01 '18 at 09:02

score 1 · Accepted Answer · answered Jan 01 '18 at 07:08

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, $\lim_{n\to\infty}a_n=0$.

Peter Szilas · Answer 2 · 2018-01-01T08:34:09.070

1

Let $n \in \mathbb{Z^+}$:

$0\lt |\dfrac{(-1)^n n^{3/2}}{5e^n}| \lt$

$\dfrac{n^2}{n^3/3!} \lt \dfrac{3!}{n}.$

And the limit is?

Used : $e^n > n^3/3!$; $n \in \mathbb{Z^+}.$

edited Jan 01 '18 at 08:34

answered Jan 01 '18 at 07:46

Peter Szilas

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score 1 · Answer 3 · answered Jan 01 '18 at 08:56

$$a_n=\frac{(-1)^n n^{\frac{3}{2}}}{5e^n} \implies b_n=|a_n|=\frac{ n^{\frac{3}{2}}}{5e^n}$$

$$\frac{b_{n+1}}{b_n}=\frac{ (n+1)^{\frac{3}{2}}}{5e^{n+1}}\frac{5e^n}{n^{\frac{3}{2}}}=\left(\frac{n+1}{n}\right)^\frac{3}{2}\frac{1}{e^n}\to1\cdot0=0$$

$$b_n=|a_n|\to0\implies a_n\to0$$

score 0 · Answer 4 · answered Jan 01 '18 at 08:45

Using Stolz Cesàro_theorem for a slightly different sequence $ \frac{ n^{2}}{5e^n}$, you'll get $\lim_{n \to \infty} \frac{ n^{2}}{5e^n}=0$ Now use the squeeze criteria to get $\lim_{n \to \infty} \frac{ n^{\frac{3}{2}}}{5e^n}= 0$. To conclude, remember that $x_n\to 0$ iff $|u_n|\to 0$

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

4 Answers4