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

3 Answers 3

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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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    @Robert: Sure, that will do it. I actually prefer a version of Gerry Myerson's approach.2012-09-06
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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 $a$?2012-09-06
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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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    Till $n_k$ the inequality is not valid, so you have to ignore first $n_k$ values of $n$2012-09-06