Thread: [SOLVED] inequalities related to binomial theorem

1. [SOLVED] inequalities related to binomial theorem

$(1+\frac{1}{n})^n\ge2, \forall n \in \mathbb{Z} : n>0$

So the easiest way I think is to use binomial theorem.

I've gotten to:

$\Sigma_{k=0}^{n-1}(_k^n)1^k(\frac{1}{n})^{n-k}+1$

So I guess I essentially have to prove:

$\Sigma_{k=0}^{n-1}(_k^n)1^k(\frac{1}{n})^{n-k}>1$

2. Originally Posted by seld
$(1+\frac{1}{n})^n\ge2$

So the easiest way I think is to use binomial theorem.

I've gotten to:

$\Sigma_{k=0}^{n-1}(_k^n)1^k(\frac{1}{n})^{n-k}+1$

So I guess I essentially have to prove:

$\Sigma_{k=0}^{n-1}(_k^n)1^k(\frac{1}{n})^{n-k}>1$
There is an alternative approach that is not hard.

Let $f(x) = \left(1+\frac{1}{x}\right)^x - 2$ for $x>0$.
Show that $f ' (x) > 0$ therefore $f$ is increasing.
At $x=2$ you can easily calculate that $f(2) > 0$.
Therefore, if $x\geq 2$ then since it is increasing it means $f(x)\geq f(2) > 0$.
But if $f(x) > 0 \implies \left(1+\tfrac{1}{x}\right)^x > 2$.
Therefore, this inequality is true for any $n\geq 2$.

3. nice, though i'm not allowed to use derivatives and such yet, we're only allowed to use inequality propositions, order axioms for R, field axioms, absolute value propositons. I'm basically taking a proofs class, and the book even says I'm supposed to use binomial theorem.

and I guess I forgot to mention it's $\forall n \in \mathbb{Z} : n > 0.$

$(1+\frac{1}{n})^n=(\frac{n+1}{n})^n\ge2$

then doing induction on the statement though I haven't made much headway.

4. Multiplying through by $n^n$, it's equivalent to showing

$(1+n)^n>2n^n$

To do that just note that

$\sum_{j=0}^n{n \choose j}n^j\geq 1 + {n \choose n-1}n^{n-1}+n^n$
$=1+n\times n^{n-1}+n^n = 1+2n^n$

That's it

5. thanks, though i feel stupid now, because that's exactly the same thing I used to prove something earlier.

6. I haven't assumed the result is true. What I have done is I've shown that $(1+n)^n>2n^n$ (you agree?). Now just divide this inequality by $n^n$ and you have the result.