Let $ a_{n} := \sqrt[n]{n},\forall n\in Z^{+}$. Show that $\lim_{n\to\infty} a_{n} = 1$
Hint: Put $a_{n}-1$ in Part(1)
We had to proof before that $(1+a)^{n}\geq$${n}\choose{2}$$a^{2},\forall n\geq2$, using the binomial theorem
My take at this was: Let $\epsilon>0, N>?, n>N$
$|a_{n} - a| <\epsilon\Leftrightarrow|\sqrt[n]{n}-1|<\epsilon$, which gives us: $(1+\epsilon)^n>n$
So now we want to show: $(1+\epsilon)^n>n$
Using: $(1+a)^{n}\geq$${n}\choose{2}$$a^{2}$ with: ${n}\choose{2}$=$\frac{n!}{2!(n-2)!} =\frac{n(n-1)}{2}$
We get:
$(1+\epsilon)^{n}\geq\frac{n(n-1)}{2}\epsilon^{2}>(n-1)\frac{\epsilon^2N}{2}$
So iff $\frac{\epsilon^2N}{2}>1$ then $\frac{\epsilon^2N}{2}>1\Leftrightarrow N>\frac{2}{\epsilon^2}$, thus we'll get:
$(n-1)\frac{\epsilon^2N}{2}>(n-1)\frac{\epsilon^2}{2}\frac{2}{\epsilon^2}=(n-1)$, so in conclusion:
$(1+a)^{n}>(n-1)$
Is this the right take? I know this should be a simpler thing to do, but we just started with this topic, and I think i missed smth while doing my steps. Any kind of feedback is appreciated ^^
Let's continue from the inequality you have obtained using binomial theorem.
Since $n\ge2$, $n-1\ge \frac{n}{2}$ $$ a^n \ge \frac{n(n-1)}{2}(a-1)^2 \ge \frac{n^2}{4} (a-1)^2 $$ substituting $a \to \sqrt[n]{n}$: $$ n \ge \frac{n^2}{4} (\sqrt[n]{n}-1)^2 \\ \sqrt[n]{n}-1\le\frac{2}{\sqrt{n}} $$ thus, for any $\epsilon>0$, there exists $N=[\frac{4}{\epsilon^2}]+1$, such that $$ n > N \ \implies|\sqrt[n]{n}-1| < \epsilon. $$