Hello

I am trying to prove that $\lim\left((2n)^{1/n}\right) = 1$.

Here $n\in \mathbb{N}$. I have already proven that $(2n)^{1/n} > 1$

for $n > 1$. So we can write $(2n)^{1/n} = 1 + k$ for some $k > 0$ when

$n>1$. Hence $2n = (1+k)^n$ for $n>1$. By the Binomial theorem, if $n>1$, we have,

\[ 2n = 1+nk+ \frac{1}{2}n(n-1)k^2+\cdots + k^n \]

Because all terms in Binomial expansion are positive, we can write

\[ 2n \geqslant 1 + \frac{1}{2}n(n-1)k^2 \]

\[ \Rightarrow \; 2n > \frac{1}{2}n(n-1)k^2 \]

\[ \Rightarrow 4 > (n-1) k^2 \]

\[ \Rightarrow k^2 < \frac{4}{n-1} \]

Above is true if $n>1$.

\[ \Rightarrow k^2 < \left(\frac{2}{\sqrt{n-1}}\right)^2 \cdots\cdots (1)\]

Now I use the following theorem I have already proven. Given $a\geqslant 0 $,

and $b\geqslant 0 $, then

\[ a<b \Longleftrightarrow a^2 < b^2 \Longleftrightarrow \sqrt{a}< \sqrt{b} \]

Now $n>1$, hence $(n-1) > 0$. Applying above theorem with $a=n-1$ and

$b=0$, we arrive at the conclusion that $\sqrt{n-1} > 0$. Hence $\frac{1}{\sqrt{n-1}} > 0$. Hence $\frac{2}{\sqrt{n-1}} > 0$.

Since $k>0$, again applying the above

stated theorem to the inequality $(1)$, we get

\[ k < \frac{2}{\sqrt{n-1}} \cdots\cdots (2)\]

for $n>1$. Now if we are given that $\varepsilon > 0$ is arbitrary, then $\frac{4}{\varepsilon^2}>0$.

\[1+\frac{4}{\varepsilon^2} > 1\]

By Archimedean Property, there exists a natural number $N_1$ such that

\[ 1< 1 + \frac{4}{\varepsilon^2} < N_1\]

\[\Rightarrow 0 < \frac{4}{\varepsilon^2} < N_1 - 1\]

\[ \Rightarrow \frac{\varepsilon^2}{4} > \frac{1}{N_1 - 1} > 0 \]

\[ \Rightarrow \left( \frac{\varepsilon}{2}\right)^2 > \left(\frac{1}{\sqrt{N_1 - 1}}\right)^2 \cdots\cdots(3)\]

Now $\frac{\varepsilon}{2} > 0$ and $N_1 >1$, so $N_1 - 1 > 0$. By above mentioned theorem, we have

$\sqrt{N_1 - 1} > 0$ , which implies $\frac{1}{\sqrt{N_1 - 1}}>0$. So applying the same theorem about the

square roots of nonnegative numbers to the equation $(3)$, we arrive at the conclusion that

\[\frac{\varepsilon}{2} > \frac{1}{\sqrt{N_1 - 1}} \]

\[ \Rightarrow \frac{2}{\sqrt{N_1 - 1}} < \varepsilon \cdots\cdots(4)\]

Now for all $n \geqslant N_1$, because $N_1 > 1$, we have $n \geqslant N_1 > 1$.

\[ \Rightarrow n-1 \geqslant N_1 - 1 > 0 \]

\[ 0 < \frac{1}{n-1} \leqslant \frac{1}{N_1 - 1} \]

Again applying the same theorem, we get

\[ \frac{1}{\sqrt{n-1}} \leqslant \frac{1}{\sqrt{N_1 - 1}} \]

\[ \Rightarrow \frac{2}{\sqrt{n-1}} \leqslant \frac{2}{\sqrt{N_1 - 1}} \]

Combining with equation $(2)$ and equation $(4)$, we conclude that, for all $n\geqslant N_1$,

$k < \varepsilon$. But since $(2n)^{1/n} = 1 + k$ , and $k > 0$, we see that

\[ | (2n)^{1/n} - 1 | = k < \varepsilon \;\;\; \forall n \geqslant N_1 \]

Since $\varepsilon$ is arbitrary, this proves that $\lim\left((2n)^{1/n}\right) = 1$.

Please let me know if proof sounds right.

thanks