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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

\begin{equation} \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}}. \end{equation}

This implies

\begin{equation} \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 \end{equation}

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?

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  • $\begingroup$ How do you knoe $\frac {2^{n^2}}{n^2} \to \infty$? $\endgroup$ – fleablood May 22 at 23:03
  • $\begingroup$ If $(a_n)\rightarrow 0$, then $(1/a_n)\rightarrow \infty$. $\endgroup$ – user2820579 May 22 at 23:05
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Without using any known limit result, consider using the inequality below: \begin{equation} \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. \end{equation}

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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

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The expression is at least

$$\frac{2^n}{\sqrt {n^4+n^4}} = \frac{1}{\sqrt 2}\frac{2^n}{n^2}.$$

Hopefully things like $2^n/n^2\to \infty$ are familiar.

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