# Proof $\frac{2^{n^2}+1}{\sqrt{n^4 + n^3}}$ diverges.

Obviously $$\lim_{n\rightarrow \infty}\frac{2^{n^2}+1}{\sqrt{n^4 + n^3}}= \infty$$, but I would like to prove this. Usually, to prove that a sequence $$(a_n)$$ diverges to $$\infty,$$ for every $$M>0$$ there is an $$n$$ such that $$a_n>M$$. In the previous example, I still can't see how to do this, so I did a workaround.

Since $$\lim_{n\rightarrow \infty}\frac{2^{n^2}}{n^2}\rightarrow \infty$$, then for $$M>0$$ there is an $$n$$ such that $$\frac{2^{n^2}}{n^2}>M$$. Thus

$$$$\frac{2^{n^2}+1}{\sqrt{n^4 + n^3}} = \frac{2^{n^2}}{n^2}\frac{1 + 1/2^{n}}{\sqrt{1+1/n}}>M\frac{1 + 1/2^{n^2}}{\sqrt{1+1/n}}.$$$$

This implies

$$$$\lim_{n\rightarrow\infty}\frac{2^{n^2}+1}{\sqrt{n^4 + n^3}} >M\lim_{n\rightarrow\infty}\frac{1 + 1/2^{n^2}}{\sqrt{1+1/n}} = M$$$$

and since the limit is bigger than every $$M>0$$, the sequence $$\frac{2^{n^2}+1}{\sqrt{n^4 + n^3}}$$ diverges.

Is this argument correct?

• How do you knoe $\frac {2^{n^2}}{n^2} \to \infty$? – fleablood May 22 at 23:03
• If $(a_n)\rightarrow 0$, then $(1/a_n)\rightarrow \infty$. – user2820579 May 22 at 23:05

Without using any known limit result, consider using the inequality below: $$$$\frac{2^{n^2} + 1}{\sqrt{n^4 + n^3}} > \frac{(1 + 1)^{n^2}}{\sqrt{2n^4}} > \frac{1 + n^2 + \binom{n^2}{2}}{\sqrt{2n^4}} > \frac{1}{\sqrt{2}} + \sqrt{2}(n^2 - 1) > n^2 > n.$$$$
You can consider $$a_n = e^{\log a_n} \geq e^{n^2 \log 2 } - e^{5\log n} \geq 2 e^{n^2 \log 2} \forall n > n_0$$ for some $$n_0$$, which clearly diverges
$$\frac{2^n}{\sqrt {n^4+n^4}} = \frac{1}{\sqrt 2}\frac{2^n}{n^2}.$$
Hopefully things like $$2^n/n^2\to \infty$$ are familiar.