Evaluating Limit Question $\lim\limits_{n\to \infty}\ \frac{1+\sqrt[2]{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}=1$?

Solution 1:

The limit immediately evaluates to $1$ after applying Stolz-Caesaro.

If one wants a more precise estimate, Euler-Maclaurin and $\sqrt[n]{n} = e^{(\log n)/n} $ gives: $$\sum_{k=1}^n \sqrt[k]{k} = \sum_{k=1}^n \left(1+ \frac{\log k}{k} + \mathcal{O}\left( \frac{\log^2 k}{k^2} \right) \right) = n + \frac{\log^2 n}{2} + \frac{\log n}{2n} + \mathcal{o}\left(\frac{\log n}{n}\right).$$

Solution 2:

The answer is $1.$ You can use the following result: If $\lim\limits_{n\to\infty }a_n=a$, then we have $$\lim_{n\to\infty} \frac{a_1+a_2+\cdots+a_n}{n}=a.$$

And you can prove this result by the definition of limits of sequences. Note that $\lim\limits_{n\to\infty}\sqrt[n]{n}=1.$

Solution 3:

For a completely elementary proof, you can use the elementary theorem(IMO) that if $a_n \to a$ then $\frac{1}{n} \sum_{k=1}^{n} a_k \to a$ as mentioned in the comments and in the other answers.

We can also do using the squeeze (but I don't see a valid proof in the comments yet).

As in my answer here: How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$?

By using $\text{AM} \ge \text{GM}$ on $k-2$ ones, $\sqrt{k}$, $\sqrt{k}$ that, for $k \ge 3$,

$$ 1 + \frac{2}{\sqrt{k}} \ge 1 - \frac{2}{k} +\frac{2}{\sqrt{k}} \ge k^{1/k} \ge 1 $$

Now we have that (using the mean value theorem, or otherwise)

$$ \frac{1}{2\sqrt{k+1}} \lt \sqrt{k+1} - \sqrt{k} \lt \frac{1}{2\sqrt{k}}$$

Adding gives us that

$$ \sum_{k=3}^{n} \frac{2}{\sqrt{k}} \le 4\sqrt{n} + 2$$

Thus

$$ n + 4\sqrt{n} + 10 \ge \sum_{k=1}^n k^{1/k} \ge n$$

And so

$$1 + \frac{4}{\sqrt{n}} + \frac{10}{n} \ge \frac{1}{n}\sum_{k=1}^n k^{1/k} \ge 1$$

and now we can apply the squeeze.