A convergence proof: $\lim_{n\to\infty} \left(1+n^2\right)^{\frac1n}$ I have to prove that
$$\lim_{n\to\infty} \left(1+n^2\right)^{\frac1n}$$
converges and give its limit.
What I did was the following:
$$\lim_{n\to\infty} \left(1+n^2\right)^\frac{1}{n} = \lim_{n\to\infty}(n^2)^{\frac1n}\lim_{n\to\infty}\left(\frac1{n^2} + 1\right)^\frac1n= \lim_{n\to\infty}(n^2)^{\frac1n}$$
Now that is where I am stuck - how to I continue with $\lim_{n\rightarrow\infty}(n^2)^{\frac{1}{n}}$?
I really don't have any ideas left...
Thank you very much!
FunkyPeanut
 A: 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 $(1+n^2)^{1/n}\to 1$.
A: 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.
A: 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, $a_n \to 0 \iff b_n \to 1$
A: Hint: $(1+n^2)^{\frac{1}{n}}=e^{\frac{\ln{(1+n^2)}}{n}}$
A: $$\left(1+n^2\right)^{\dfrac1n}=\left(1+\dfrac1{n^2}\right)^{\dfrac1n}\cdot n^{\dfrac2n}=\left(\left(1+\dfrac1{n^2}\right)^{n^2}\right)^\dfrac1{n^3}\cdot\left(n^{\dfrac1n}\right)^2$$
Now we know $\displaystyle\lim_{h\to0}\left(1+h\right)^{\dfrac1h}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$
So as $n\to\infty,$ the first limit reduces to $e^{\frac1{\infty}}=\cdots$
and we can prove $\displaystyle n^{\frac1n}=1$
