Computation of the succession $\lim_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right)$

Question

For training I have decided to solve this limit of a succession

$$\lim\limits_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right).$$

My first attempt:

\begin{split} \lim_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right) =\\ &=\lim_{n\to \infty } 3^{n+1}\left(1-\frac{3^{n\cdot \sqrt{1-\frac{1}{n^2}}}}{3\cdot 3^{n}}\right)\\ &=\ldots \end{split}

I have abandoned this approach because I wanted to use a notable limit (if $\{b_n\}\to 0$) that could probably be useful; but seeing the rounded brackets I have thought that I have occurred many times:

$$\lim_{n\to \infty} \frac{a^{b_n}-1}{b_n}$$

My second attempt:

\begin{split} \lim_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right) =\\ &=\lim_{n\to \infty } 3^{\sqrt{n^2-1}}\left(\frac{3^{n+1}}{3^{\sqrt{n^2-1}}}-1\right)\\ &=\ldots \end{split}

Considering only

$$\frac{3^{n+1}}{3^{\sqrt{n^2-1}}}=3^{(n+1-\sqrt{n^2-1})}$$

Taking only the exponent I have:

$$n+1-\sqrt{n^2-1}=\frac{((n+1)-\sqrt{n^2-1})\cdot ((n+1)+\sqrt{n^2-1})}{(n+1)+\sqrt{n^2-1}}$$ $$=\frac{2n+2}{n+1+\sqrt{n^2+1}} \tag 1$$

If I take the limit of the $(1)$, easily:

$$\lim_{n\to \infty}\frac{2n+2}{n+1+\sqrt{n^2+1}}=1$$

i.e.

$$\lim_{n\to \infty}3^{(n+1-\sqrt{n^2-1})}=3$$

Definitely I think that this limit is $\infty\cdot 2=+\infty$. I think that is correct. Do you users think there are better alternatives? Thanks in advance.

Answer 1

For $n>1$ you have $$3^{n+1}-3^{\sqrt{n^2-1}}>3^{n+1}-3^{\sqrt{n^2}}=3^{n+1}-3^n=2\cdot3^n.$$

Answer 2

The obvious way to obtain $\infty$ as the limit:

$$ 3^{n+1} - 3^{\sqrt{n^2 - 1}} \ge 3^{n+1} - 3^{n} = 2(3^n) \rightarrow \infty. $$ Your second attempt seems correct but is more complicated.