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.


2 Answers 2


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


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.

  • $\begingroup$ How can you prove that both of the sequences tends to 0 as n tends to infinity? $\endgroup$
    – Lucy
    Nov 1, 2014 at 15:29
  • $\begingroup$ 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. $\endgroup$
    – mookid
    Nov 1, 2014 at 15:32
  • $\begingroup$ and $2^{3n}=8^n$ to finish. Cheers :-) $\endgroup$ Nov 1, 2014 at 15:41
  • $\begingroup$ @mookid Thank you for the details. $\endgroup$
    – Lucy
    Nov 1, 2014 at 15:55

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


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