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I want to show that $(1+ \frac{1}{k})^k \geq 2$ for say $k \geq 2$. Here is what I have so far:

By the Binomial Theorem we know $(1 + x)^k$ for $k \geq 2$ gives us:

$1^k + {k\choose 1}1^{k-1}x^1 + {k\choose 2}1^{k-2}x^2$ which yields:

$1 + kx + \frac{k(k-1)}{2}x^2$

We know substitue $x = \frac{1}{k}$:

$1 + 1 + \frac{1}{2} - \frac{1}{k}$

Back to the inequality we care about:

$\frac{5}{2} - \frac{1}{k} \geq 2$

Subtracting $2$ from both sides:

$\frac{1}{2} - \frac{1}{k} \geq 0$

Thus, this is true for all $k \geq 2$

Is this correct?

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The binomial theorem gives you more than those three terms in general. – Marc van Leeuwen Sep 21 '12 at 6:52
up vote 3 down vote accepted

The Binomial Theorem says that if $k$ is a positive integer, then $$(1+x)^k=1+\binom{k}{1}x+\binom{k}{2}x^2+\cdots +\binom{k}{k}x^k.$$ Note that if $x$ is positive, then each term in the binomial expansion above is positive.

So for $k\ge 2$, we have $$(1+x)^k\ge 1+kx+\binom{k}{2}x^2.$$

In particular, if we put $x=1/k$, we find that $$\left(1+\frac{1}{k}\right)^k\gt 1+k\cdot \frac{1}{k}=2.$$

Remark: The fact that $(1+x)^n \ge 1+nx$ for $x$ positive is called Bernoulli's Inequality. (It can be stretched to $x\ge -1$, but that is irrelevant for us.)

The Bernoulli Inequality can also be proved by a simple induction argument, we do not really need the Binomial Theorem. For if we know that $(1+x)^n \ge 1+nx$, then $(1+x)^{n+1}=(1+x)^n(1+x)\ge (1+nx)(1+x)=1+(n+1)x+nx^2\ge 1+(n+1)x$.

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@enzotib: Thanks, I think I found it. It will take me another couple of read-throughs to fix nearly all the typos. – André Nicolas Sep 21 '12 at 5:55
I will prefer to use common theorems. Anyway how could I show that $(1 + \frac{1}{k})^k \leq 3$ – CodeKingPlusPlus Sep 21 '12 at 5:56
$(1+\frac{1}{k})^k$ is an increasing function and $\lim_{k\to \infty} (1+\frac{1}{k})^k=e\implies (1+\frac{1}{k})^k\lt 3$, even equality doesn't hold. – Aang Sep 21 '12 at 6:04

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