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I need to prove without using L'Hopital rule that $$\lim_{n\to\infty}{\frac{2^n}{n}}=\infty \; n\in\mathbb N$$ and $$\lim_{n\to\infty}{\frac{3^n}{n}}=\infty$$ What's the best way?

leonbloy
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Breldor
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  • Not quite sure what you mean there, as far as I can tell $\lim a_n =\infty$ is a rather common way to write this kind of limit? Instead of going all out with "show that for each $K\in\mathbb{R}$ there is a $n_0$, such that..." - and in that case, those expressions are perfectly sensible, unless I'm missing something important here? – Some Math Student Nov 09 '14 at 22:45

4 Answers4

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Let $n\ge 2$. By the Binomial Theorem, $(1+1)^n\ge 1+n+\frac{n(n-1)}{2}\gt \frac{n(n-1)}{2}$.

André Nicolas
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Using $\frac{(n+1)^2}{n^2}=1+\frac2n+\frac1{n^2}<2$ for $n>2$, show by induction that $2^n>n^2$ for all $n\ge3$

Hagen von Eitzen
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Try to show that for every $t\in\mathbb{N}$ there is a $n_0$, such that $\frac{2^n}{n}>t\quad\forall n>n_0$. You could also show something along the lines $\frac{2^n}{n}>2^{\frac{n}{2}}$ or so, getting rid of your denominator. For the second sequence, think about $2^n<3^n$ and what it means for you.

Some Math Student
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If you use the basic fact that for $n\ge5$ $$2^n>n^2$$ you can show that definitively $$\lim_{n\to \infty}\frac{2^n}{n}>\lim_{n\to \infty}\frac{n^2}{n}=\lim_{n\to \infty}n=\infty$$

Generally speaking, for every $a \in (1,+\infty), b\in \mathbb{R}$ there always exists an $N\in \mathbb{N}$ such that for every $n\geq N$ you have

$$a^n>n^b$$

Intuitively, this is why exponentials always "prevail" on powers.

Lonidard
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