# Proof of a binomial theorem based inequality?

Let $k \in N, x \gt 0$. Show that there exists some $n_2 \in \mathbb{N}$ so that $\forall n \geq n_2: (1+x)^n \gt n^k$. Hint: binomial theorem.

My thought on this is first to make the substitution $(1+x)=b$ which means $b>1$ and $b^n>1$. This would also be true if $k=0$ and $n=1$ thus $n_2=1$.

Next step I can think of is using archimedian property $a>0, y \in \mathbb{R}, m \in \mathbb{N}, ma>y$. This will result in $b^n=(1+x)^n=ma > y = n^k$. My current idea would be to replace $n$ with some other cleverly devised number OR using induction because of the request for all $n$ and maybe then I can use binomial theorem to finally solve this. What's really throwing me off is that $n$ is the exponent on the left and also the base on the right (that is why I thought about replacing $n$).

Any hints on how to get to the next step? Thanks.

-

My first thought was to assume $n\ge k+1$ for starters. Then we can leave out all but one term from the binomial formula to get a very crude estimate $$(1+x)^n>x^{k+1}\binom{n}{k+1}.$$ Then divide that with $n^k$ and take the limit as $n\to\infty$. We get $$\lim_{n\to\infty}\frac{x^{k+1}n(n-1)\cdots (n-k)}{(k+1)!n^k}=\infty,$$ because upstairs we have a degree $k+1$ polynomial of $n$ and downstairs we only have a degree $k$ polynomial. Surely then this ratio is $>1$ from some point on :-)

-
I'm a relatively new guy here. Is there an established site policy on homework threads? If you think that I am giving out too much detail/hints, please say so! I certainly don't want to rock the boat. Downvote, if necessary :-) – Jyrki Lahtonen Jun 8 '11 at 20:34

I don't know how to use the binomial theorem in this problem. But here is my take. One can show that $\ln z/z$ is decreasing for $z>e$ and approaching $0$. Hence there is an $n_2$ such that $$\frac{\ln n}{n}\leq\frac{\ln n_2}{n_2}<\frac{\ln(1+x)}{k}$$ for $n\geq n_2$.

It follows that $$k\ln n < n\ln(1+x)\Leftrightarrow n^k < (1+x)^n$$

-

Here is an idea which might work: Pick an $l$ so that $lx \geq k!$ (you might need some other expression here) and then

$$(1+x)^{ml} = ((1+x)^l)^m \,.$$

By Bernoulli

$$(1+x)^l \geq 1+lx \,.$$

Thus

$$(1+x)^{ml} \geq (1+lx)^m \,.$$

When $m \gg k$ the term $\frac{m!}{k! (m-k)!} (lx)^m$ is a term in the sum, and this looks to me much larger than $m^kl^k$.

P.S. Just to clarify: I think that the original inequality is easy to prove if $x \geq 1$. The trick at the begining tries to reduce the inequality to something of this type, but you probably need to figure the right right side.

Also this inequality is easy to prove with Analysis/Calculus, but I guess that's not acceptable in this class.

-