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Does $\displaystyle\lim_{n\to \infty}\ \frac{1+\sqrt[2]{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}=1$?

Thanks a lot for your time and help.

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Perhaps this. – David Mitra Apr 11 '12 at 13:45
The sequence n^1/n decreases for n >=3.Your sequence is bigger than n times 1/n since it is the minimal sumand.And this sequence converges to 1. Try to find a bigger sequence that converges also to 1. – alpha.Debi Apr 11 '12 at 13:58
@Anonymous No, the result follows immediately from the theorem in the link (take $a_n=\root n\of n$). But this may be overkill; see alpha's comment. – David Mitra Apr 11 '12 at 14:05
@alpha.Debi: Please make sure to put in parentheses. n^1/n=1 because the exponentiation takes precedence. You are right that n^(1/n) decreases with n starting with 3. – Ross Millikan Apr 11 '12 at 14:09
THREE answers have been posted, none of them by me, but so far the ONLY up-vote for this question is mine! – Michael Hardy Apr 11 '12 at 15:45
up vote 1 down vote accepted

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 $\lim\limits_{n\to\infty}\sqrt[n]{n}=1.$

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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: $$\sum_{k=1}^n \sqrt[k]{k} = \sum_{k=1}^n \left(1+ \frac{\log k}{k} + \mathcal{O}\left( \frac{\log^2 k}{k^2} \right) \right) = n + \frac{\log^2 n}{2} + \frac{\log n}{2n} + \mathcal{o}\left(\frac{\log n}{n}\right).$$

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Beautiful answer! – Tyler Apr 11 '12 at 14:19
Good answer! But I think it is difficult for a beginner! – Riemann Apr 11 '12 at 14:51
@Riemann I agree my answer is perhaps not ideal for a beginner, the squeeze method mentioned in the comments above is probably the simplest. I just thought it would be interesting for the OP and others to see other approaches. – Ragib Zaman Apr 11 '12 at 14:55
@Ragib Zaman Yes, of course. I quite agree with you. – Riemann Apr 11 '12 at 15:08
I don't see any valid proof by the squeeze method in the comments. – Aryabhata Apr 11 '12 at 15:14

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: $\lim_{n \to +\infty} n^{\frac{1}{n}} $

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

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


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

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.

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If you don't want to use mean value theorem, $\sqrt{k+1} - \sqrt{k} = \frac{1}{\sqrt{k+1} + \sqrt{k}} \lt \frac{1}{\sqrt{k} + \sqrt{k}}$ – Aryabhata Apr 11 '12 at 15:35
Thanks, I'm trying to figure out your proof. why does $\frac{1}{2\sqrt{k+1}} \lt \sqrt{k+1} - \sqrt{k}$? – Anonymous Apr 11 '12 at 15:59
@Anonymous: You don't really need that part, but $\frac{1}{\sqrt{k+1} + \sqrt{k}} \gt \frac{1}{\sqrt{k+1} + \sqrt{k+1}}$ – Aryabhata Apr 11 '12 at 16:15
OK, that part I understood, but I don't see how it relates to your answer. – Anonymous Apr 11 '12 at 16:25
@Anonymous: It does not. It might make it easier to see how the mean value theorem applies though. – Aryabhata Apr 11 '12 at 16:39

Set $a_n=1+\root 2\of 2+\root3\of 3+\cdots +\root n\of n$.

Here is a proof based on the facts that

$\ \ \ \ \ $1) $\lim\limits_{n\rightarrow\infty} \root n\of n=1$


$\ \ \ \ \ \ $2) $(\root n\of n)$ is decreasing for $n\ge3$.

Trivially, ${a_n\over n}\ge 1$ for all $n$; thus, $$\tag{1} \liminf_{n\rightarrow\infty}{a_n\over n}\ge 1. $$

By 2), we have for a fixed $M\ge 3$ and $k$ a positive integer:

$$\eqalign{ {a_{M+k}\over M+k} \ \ &\ \ ={a_M\over M+k}+{ (M+1)^{1/M+1} + (M+2)^{1/M+2}+\cdots+(M+k)^{M+k}\over M+k }\cr \ \ &\ \ \le {a_M\over M+k}+{ k(M+1)^{1/M+1} \over M+k}\cr & \buildrel{ k\rightarrow\infty}\over\longrightarrow \ \ 0+{(M+1)^{M+1}}.\phantom{M\over M} } $$ From this it follows that $\limsup\limits_{n\rightarrow\infty} {a_n\over n}\le (M+1)^{1/M+1}$ for all $M\ge3$; and, thus by 1), we have $\limsup\limits_{n\rightarrow\infty} {a_n\over n}\le1$. Combining this result with $(1)$, we have $\lim\limits_{n\rightarrow\infty }{a_n\over n}=1$.

This is just a modification of the proof of the Cesaro theorem found here. The decreasing condition 2) used in the above is not needed, and, alternatively, you could just use the proof in the link.

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