Using sandwich theorem, I can prove that


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?


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.


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}} $.


We have

$$ \int_1^y x^y dx = \frac{y^{y+1}-1}{y+1} $$


$$ \frac{1}{n}\left(\sum_{k=1}^n k^n\right)^{\frac{1}{n}}\le \frac{1}{n}\left( \frac{n^{n+1}-1}{n+1}\right)^{\frac{1}{n}} \le \frac{1}{n}\left( \frac{n^{n+1}}{n+1}\right)^{\frac{1}{n}} $$


$$ \lim_{n\to\infty}\frac{1}{n}\left( \frac{n^{n+1}}{n+1}\right)^{\frac{1}{n}} = 1 $$

The left side limit is left as an exercise.


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