Prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals. I need to prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals.
I started by using Bernoulli's inequality: 
$(1+\frac{2}{\sqrt{n}}) < (1 + \frac{1}{\sqrt{n}})^2$
I can say that:
$(1+\frac{2}{\sqrt{n}}) = (1+\frac{2\sqrt{n}}{n})$
I can also subtract the one and divide by 2 on the left side without changing the inequality (because it makes it even smaller): $(\frac{\sqrt{n}}{n}) < (1 + \frac{1}{\sqrt{n}})^2$
But now I am stuck...

Well for natural (positive) $n$ then  $\sqrt[n]{n} \le (1+ \frac 1{\sqrt n})^2 \iff$
$n \le (1 + \frac 1{\sqrt n})^{2n}$
So $(1+\frac 1{\sqrt n})^{2n} \ge 1 + \frac {2n}{\sqrt n}$ and dang... that's not enough.
But lets go one more term.
Remember the reason $(1 + b)^n \ge 1+ nb$ is becase $(1 + b)^n = 1 + nb + C_2b^2 + C_3b^3 + ..... + b^n$ and $C_kb^k \ge 0$.  (Note if $b$ is large and $n>0$, in particular if $b \ge 1$, this is $(1+b)^n \ge 1 + nb$ is a very large discrepancy in the inequality).
So $(1+\frac 1{\sqrt n})^{2n} \ge 1 + \frac {2n}{\sqrt n} + {2n \choose 2}\frac 1{n}$.
So ${2n \choose 2} = \frac {2n(2n-1)}2 = n(2n-1)$ so
$(1 + \frac 1{\sqrt n})^{2n} \ge 1 +\frac{2n}{\sqrt n} + n(2n-1)\frac 1n =$
$1 + 2\sqrt n + 2n- 1 = 2(\sqrt n + n) > n$.
.... which if we did that correctly is not even close.
