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I'm interested in the proofs of the following: Using the definition of limit (for sequences).

I proved the first one, as shown, but I don't know how to go about doing the second and third.

  1. $\displaystyle \lim_{n \to \infty} \frac{n}{3^n} = 0$. Proof: Let $\epsilon > 0$. If $n > M$, then $\vert \frac{n}{3^n} - 0\vert = \frac{\displaystyle n}{\displaystyle \Big(\frac32\Big)^n2^n} \leq \frac{\displaystyle n}{\displaystyle (1 + \frac{n}2)(1 + n)} \leq \frac{\displaystyle n}{\displaystyle (1 + n/2)n} = \frac{1}{\displaystyle 1 + \frac12 n} \leq \frac{1}{\frac12 n} = \frac2n < \epsilon$ when $\displaystyle M = \frac{2}{\epsilon}$.

But for these I can't seem to get them. Please provide a proof. Or a good hint! Thanks!

  1. $\displaystyle \lim_{n \to \infty} \frac{n^3}{2^n} = 0$.

  2. $\displaystyle \lim_{n \to \infty} \frac{n^6}{3^n} = 0$.

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up vote 4 down vote accepted

For the second problem, suppose that $n \ge 4$. Then by the Binomial Theorem, $$2^n=(1+1)^n=1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}+\frac{n(n-1)(n-2)(n-3)}{24}+\cdots.$$ In particular, $$2^n\gt \frac{n(n-1)(n-2)(n-3)}{24}.$$

If $n \ge 6$, then $n-1\gt n/2$, $n-2\gt n/2$, and $n-3\ge n/2$. So $$\frac{n^3}{2^n}\lt \frac{192}{n}.$$

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For the second one I would show $3^n > n(n-1)(n-2)(n-3)(n-4)(n-5)\cdot 2^7/6!$ by the Binomial Theorem and then each $n,n-1, \cdots, n-5 > n/2$ for $n \geq 10$? – Robert Cardona Sep 6 '12 at 7:40
@Robert: Sure, that will do it. I actually prefer a version of Gerry Myerson's approach. – André Nicolas Sep 6 '12 at 9:46

If you can do $n/3^n$, I bet you can do $n/a^n$ for any $a\gt1$. Then note that $n^3/2^n=(n/a^n)^3$ for some cleverly chosen $a$, and similarly for $n^6/3^n$.

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For some cleverly chosen $n$, or some cleverly chosen $a$? – Robert Cardona Sep 6 '12 at 7:46
Thanks, Robert - edited. – Gerry Myerson Sep 6 '12 at 13:20

Hint: Can you prove by induction that $$\frac{n^k}{3^n}<\frac{1}{n}\hspace{5 mm} i.e. \hspace{5mm} n^{k+1}<3^n$$ after some finitely many (depending on $k$) values of $n$, say $1,2,\ldots n_k$ ? Then you can choose any $M>\max {\lbrace 1/\epsilon,n_k+1}\rbrace$

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This approach would actually work for both (2) and (3) right? Except for 2 it would be showing $n^{k+1} < 2^n$ after some value of $n$ – Robert Cardona Sep 6 '12 at 7:26
@Robert: Absolutely – pritam Sep 6 '12 at 7:27
What is $n_k + 1$ represent? – Robert Cardona Sep 6 '12 at 7:49
Till $n_k$ the inequality is not valid, so you have to ignore first $n_k$ values of $n$ – pritam Sep 6 '12 at 7:52

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