Without squeeze theorem Using sandwich theorem, I can prove that
$\lim_{n\to\infty}\Big((\frac{1}{n})^n+(\frac{2}{n})^n+...+(\frac{n}{n})^n\Big)^{\frac{1}{n}}=1$
But I am curious, is there any way to solve it without the squeeze theorem? If the answer is positive, could you help me to find it?
 A: For any $n\in\mathbb{N}$ the sum
$$ \sum_{k=1}^{N} k^n $$
is a polynomial in $N$ with degree $n+1$ and leading term $\frac{N^{n+1}}{n+1}$, as a consequence of the hockey stick identity. In particular
$$ \sum_{k=1}^{n}\left(\frac{k}{n}\right)^n =\frac{n}{n+1}+o\left(1\right) $$
as $n\to +\infty$, which can be proved through Riemann sums too. The claim is a straightforward consequence.
A: To show a limit
involving a sum of this type,
squeezing would usually be involved.
Here is a derivation of
explicit bounds.
Since
$x^n$ is an increasing function
if $x > 0$ and $n > 0$,
$(k/n)^n
\lt \int_{k/n}^{(k+1)/n} t^n dt
\lt ((k+1)/n)^n
$.
Summing,
$\sum_{k=0}^{n-1} (k/n)^n
\lt \sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n} t^n dt
\lt \sum_{k=0}^{n-1}((k+1)/n)^n
$
or
$\sum_{k=0}^{n-1} (k/n)^n
\lt \int_0^1 t^n dt
=\dfrac1{n+1}
\lt \sum_{k=1}^{n}(k/n)^n
$
or
$ \dfrac1{n+1}
\lt \sum_{k=1}^{n}(k/n)^n
\lt 1+\dfrac1{n+1}
$.
Therefore
$ \dfrac1{(n+1)^{1/n}}
\lt \left(\sum_{k=1}^{n}(k/n)^n\right)^{1/n}
\lt (1+\dfrac1{n+1})^{1/n}
$.
Since
$(1+x)^n
\ge 1+nx$,
$(1+x/n)^n
\ge 1+x$
so
$(1+x)^{1/n}
\le 1+x/n$.
Therefore
$(1+\dfrac1{n+1})^{1/n}
\le 1+\frac1{n(n+1)}
$.
Since
$(1+\frac1{\sqrt{n+1}})^n
\ge 1+\sqrt{n+1}
\gt \sqrt{n+1}
$,
raising to the
$2/n$ power,
$(n+1)^{1/n}
\lt (1+\frac1{\sqrt{n+1}})^2
\lt (1+\frac{2}{\sqrt{n+1}}+\frac1{n+1})
\lt (1+\frac{3}{\sqrt{n+1}})
$.
Therefore
$\frac1{(n+1)^{1/n}}
\gt \frac1{(1+\frac{3}{\sqrt{n+1}})}
= \frac{\sqrt{n+1}}{3+\sqrt{n+1}}
= 1-\frac{3}{3+\sqrt{n+1}}
$.
A: We have
$$
\int_1^y x^y dx = \frac{y^{y+1}-1}{y+1}
$$
then 
$$
\frac{1}{n}\left(\sum_{k=1}^n k^n\right)^{\frac{1}{n}}\le \frac{1}{n}\left( \frac{n^{n+1}-1}{n+1}\right)^{\frac{1}{n}} \le \frac{1}{n}\left( \frac{n^{n+1}}{n+1}\right)^{\frac{1}{n}}
$$
and
$$
\lim_{n\to\infty}\frac{1}{n}\left( \frac{n^{n+1}}{n+1}\right)^{\frac{1}{n}} = 1
$$
The left side limit is left as an exercise.
