# Finding the limit of a given sequence: ${a_{n}} = \frac {2^{3n}-n3^n} {n^{1729}+8^n}$

I have a limit as follows

$${a_{n}} = \frac {2^{3n}-n3^n} {n^{1729}+8^n}$$

and I would like to find the limit of this sequence.

I think the limit is 1, so I tried to prove this using the squeeze rule and I am not able to do it.

Thanks to anybody who helps.

Hint as $$\frac{n3^n}{8^n}\to 0 \\ \frac{n^{1729}}{8^n}\to 0$$ the limit is $1$.

Details:

let us prove that $\frac{n3^n}{8^n} \le C\frac1{\sqrt n}$ for a certain $C$. We use induction:

• if $\frac{n3^n}{8^n} \le C\frac1{\sqrt n}$ then $$\frac{(n+1)3^{n+1}}{8^{n+1}} = \frac{n3^n}{8^n} \times \frac 38\frac{n+1}n \le C\frac1{\sqrt n} \times \frac 38\frac{n+1}n \le C \frac1{\sqrt {n+1}}$$ The last inequality is true is soon as: $$C\frac 38 \left(\frac{n+1}n\right)^{3/2} \le 1 \\ \iff n\ge \frac 1{ \left(\frac 8{3C}\right)^{2/3} - 1}$$ which is always true if $C$ is large enough.

• now choose $C$ so that the inequality is true for $n=1$ as well. And you are done.

• How can you prove that both of the sequences tends to 0 as n tends to infinity?
– Lucy
Nov 1, 2014 at 15:29
• in both cases you can prove using induction that it is dominated by, say, $C/\sqrt n$ for a certain $C$. I can add details if you like. Nov 1, 2014 at 15:32
• and $2^{3n}=8^n$ to finish. Cheers :-) Nov 1, 2014 at 15:41
• @mookid Thank you for the details.
– Lucy
Nov 1, 2014 at 15:55

Asymptotically the exponential terms carry, i.e. $$2^{3n} / 8^n = 1$$ easily.