# Some algebraic inequalities with the binomial theorem.

I am working on proving the following limits.

1), $$\lim_{n \to \infty} \sqrt[n]{n} = 1$$

2), If $$p >0$$ and $$\alpha \in \Bbb R$$, then $$\lim_{n \to \infty} {n^{\alpha}\over{(1+p)^n}} =0$$

And I am trying to follow the proof but I am stuck on understanding the following inequalities regarding the binomial theorem.

Put $$x_n = \sqrt[n]{n} -1$$. then $$n = (1+x_n)^n \ge {n(n-1) \over 2}x_n^2$$

I understand that the right side of the inequality probably comes from $$1 + nx_n + {n \choose 2}x_n^2 + ・・・+x_n^n$$

and taking all the terms after $${n \choose 2}x_n^2$$.

But to know this I think that $$x_n^k$$ has to be greater than $$x_n^2$$ for $$k > 2$$, am I right? Since we can only say that $$x_n > 0$$, I'm a little puzzled.

For the second one,$$(1+p)^n > {n \choose k}p^k = {n(n-1)・・・(n-k+1) \over {k!}}p^k > {n^kp^k \over {2^k k!}}$$

is supposedly true when $$n > 2k$$.

I understand the left three expressions, but how does the right side of the inequality work ?

The first confusion is just a misunderstanding. If you have $A=B+C$ with $B,C$ greater than $0$, then no matter what happens to $B,C$'s relationship, $A>B$ is always true.

For the second one all you need to show is $(n-k+1)\ge \frac{n}{2}$ if $n\ge 2k$. But this should be clear to you via some algebraic manipulations.

• Ah, I see. It's really tough to self teach analysis. But thanks ! Now I can move on to the next theorems. Commented May 17, 2013 at 7:59

For the first one: since it is true that $\,\sqrt[n] n\ge 1\;\;\forall\,n\in\Bbb N\,$ (why?) , we can write

$$\sqrt[n] n=1+c_n\;,\;\;0\le c_n\in \Bbb R\implies$$

$$\implies n=(1+c_n)^n=\sum_{k=0}^n\binom nkc^k_n\ge\binom n2c_n^2=\frac{n(n-1)}2c_n^2\implies$$

$$\implies 0\le c_n\le\sqrt{\frac2{n-1}}\xrightarrow[n\to\infty]{}0$$

The squeeze theorem then gives

$$c_n\xrightarrow[n\to\infty]{}0\implies \sqrt[n]n=1+c_n\xrightarrow[n\to\infty]{}1$$

• Ah. Thanks. I actually only needed to know why (1+c_n)^n > n(n-1)/2 c_n^2 because I already knew the rest of the proof, but since you put some extra steps inside it actually made sense. Commented May 17, 2013 at 7:57