A convergence proof: $\lim_{n\to\infty} \left(1+n^2\right)^{\frac1n}$

Solution 1:

Define a sequence $\{a_n\}_{n\in\mathbb N}$ as $$ (1+n^2)^{1/n}=1+a_n, $$ then $a_n>0$, and $$ 1+n^2=(1+a_n)^n\ge 1+na_n+\frac{n(n-1)}{2}a_n^2+\frac{n(n-1)(n-2)}{6}a_n^3\ge \frac{n(n-1)(n-2)}{6}a_n^3. $$ Thus $$ a_n^3\le \frac{6n^2}{n(n-1)(n-2)}=\frac{6n}{(n-1)(n-2)}, $$ and hence $$ 0\le a_n\le \left(\frac{6n}{(n-1)(n-2)}\right)^{1/3}\to 0, $$ and thus $a_n\to 0$ and hence $(1+n^2)^{1/n}\to 1$.

Solution 2:

One of many ways: take log of the expression and show that $$ \log L = \lim_{n \to \infty}\frac{2 \log n}{n}=0 $$ Hence $L=e^0=1$

EDIT: If for some reason you don't quite like using $\log(1+n^2) \sim 2 \log n $ for large $n$, rewrite it as $\log (n^2(1+\frac{1}{n^2})=2\log n +\log(1+\frac{1}{n^2})$ and take the limit of the second term, which is of course $0$. These two approaches are essentially the same.

Solution 3:

This is similar to Yiorgos' proof, but I'm going to follow the comments and show the proof that $n^{1/n} \to 1$

Claim: $n^{1/n} \to 1$.

Call $b_n := n^{1/n}, a_n := b_n - 1$ then we have that $$ n = (a_n + 1)^n = \sum_{i=0}^n \binom{n}{i} a_n^i 1^{n-i} \ge \binom{n}{0} + \binom{n}{1}a_n + \binom{n}{2}a_n^2 = 1 + na_n + \frac{n!}{2!(n-2)!}a_n^2 $$ thus we can simplify the above expression to $$ n \ge 1 + na_n + \frac{n(n-1)}{2}a_n^2 \ge \frac{n(n-1)}{2}a_n^2 \iff a_n \le \frac{2}{n-1} $$ Place $c_n = \frac{2}{n-1}$ then

$$ c_n = \frac{\frac{2}{n}}{1-\frac{1}{n}} \to \frac{0}{1-0} = 0 \text{ as } n \to \infty $$ by the linearity of the limit. Now we need to take a quick break from the immediate problem to prove $b_n \ge 1 \; \forall n \in \mathbb{N}$:

We know that $$ n \ge 1 \iff n \ge 1^{n} \iff n^{1/n} \ge 1 \iff b_n \ge 1 \; \forall n \in \mathbb{N} $$

so place $d_n := 0$ (clearly $d_n \to 0$ as $n \to \infty$) then we get the following inequality: $$ d_n \le a_n \le c_n $$ whence, by squeeze theorem, $a_n \to 0 \iff b_n \to 1$