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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3Perhaps this. – David Mitra Apr 11 '12 at 13:45
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3The 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
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@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
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1@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
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OK, I was talking about n^(1/n), the general sumand of the numerator of his question. – alpha.Debi Apr 11 '12 at 14:14
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@alpha.Debi could you please format your comment above stating: "you can express the sequence as the summ of the two first sumands with the other, and the n-3 other are lower than n-3 times 3^1/3 / n that converges to 3^1/3" in TeX so I can understand which sequence you mean? Thanks a lot. – Anonymous Apr 11 '12 at 14:20
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See the explanation of David Mitra.What I said is no necessary:AL your sequence is lower than n times (the third sumand) /n.But it does not help because you should find a bigger sequence that converges to 1.That was to explain that the final limit is finit , because you wrote infinite , now I see youwrote 1. – alpha.Debi Apr 11 '12 at 14:32
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@David Mitra isn't the limit of $n\cdot 3^{1/n}=\infty$ because $\lim \limits_{n\to \infty}\ 3^{1/n}=1$ and $\lim \limits_{n\to \infty}\ n=\infty$? – Anonymous Apr 11 '12 at 14:32
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@David Mitra how do I show that $\sqrt[k]{k}\le 3^{1\over n}$ for every $1\le k\le n$? (BTW, we haven't gotten to derivations in the course yet). – Anonymous Apr 11 '12 at 14:57
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in the $n$ numbers $1,\sqrt[2]{2},\sqrt[3]{3},...,\sqrt[n]{n}$ the biggest one is $\sqrt[3]{3}$. – Riemann Apr 11 '12 at 15:11
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Yes, blind spot, deleted. – André Nicolas Apr 11 '12 at 15:19
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@Aryabhata Yes, of course. Complete brain freeze on my part. Deleting... – David Mitra Apr 11 '12 at 15:26
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@David Mitra did you mean $\sqrt[k]{k}\le 3^{1\over n}$ or $\sqrt[k]{k}\le 3^{1\over 3}$? if you meant $3^{1\over 3}$ I can't use the squeeze therorem with that as it doesn't converges to 1. – Anonymous Apr 11 '12 at 15:30
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1THREE 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
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@MichaelHardy: I would have, if it had shown some prior effort. – Aryabhata Apr 11 '12 at 16:16
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@Aryabhata I'm a new student in Math, unfortunately I'm not too advanced as you can see from my questions. However, I'm doing as much as I can before asking questions here, although sometimes it seems to more advanced people that no efforts have been made. – Anonymous Apr 11 '12 at 16:24
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@Anonymous: You can always mention what approaches you took. – Aryabhata Apr 11 '12 at 16:26
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@Aryabhata in the original post you can see my initial attempt, as this post was edited several times. – Anonymous Apr 11 '12 at 16:28
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@Anonymous: I see. +1 then. In future, I suggest you leave your working in. – Aryabhata Apr 11 '12 at 16:30
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@Aryabhata I'll certainly keep that in mind for the next times. Thank you very much for your time and efforts. – Anonymous Apr 11 '12 at 16:31
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By considering $f(x)=\dfrac{1}{x^x}$, and the partitions ${0,1/n,\ldots (n-1)/n,1}$ of $[0,1]$, I was tempted to say that $$\frac{1+\sqrt[2]{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}=\sum_{k=1}^n \frac{1}{n}f(1/k),$$ is a Riemann sum of $f$ over $[0,1]$, so $$\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{n}\sqrt[n]{k}=\int_0^1 \frac{1}{x^x}dx,$$ but it seems that $f$ don't have primitives in terms of elementary functions. – leo Apr 11 '12 at 23:04
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See also: Computing $\lim_{n\rightarrow \infty} \frac{1+\sqrt{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}$. – Martin Sleziak Apr 17 '18 at 07:13
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Since several answers mentioned this result, it's probably good to add a link to a post where it is also proved: On Cesàro convergence: If $ x_n \to x $ then $z_n = \frac{x_1 + \dots +x_n}{n} \to x$. – Martin Sleziak Dec 13 '19 at 13:31
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@leo just wanted to point out that your comment here is a bit misleading (I stumbled upon it and it really confused me first). Even if your $f$ did have an antiderivative, it wouldn't help because the limit you have in the LHS is not the one required in the question. – Sayan Dutta Jun 04 '23 at 19:50
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@SayanDutta it is, it's just I made a mistake. It should be $\sqrt[k]{k}$ instead of $\sqrt[n]{k}$. – leo Jun 06 '23 at 03:20
4 Answers
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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: $$\sum_{k=1}^n \sqrt[k]{k} = \sum_{k=1}^n \left(1+ \frac{\log k}{k} + \mathcal{O}\left( \frac{\log^2 k}{k^2} \right) \right) = n + \frac{\log^2 n}{2} + \frac{\log n}{2n} + \mathcal{o}\left(\frac{\log n}{n}\right).$$
Ragib Zaman
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@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
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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 $\lim\limits_{n\to\infty}\sqrt[n]{n}=1.$
Riemann
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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.
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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
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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
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@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
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OK, that part I understood, but I don't see how it relates to your answer. – Anonymous Apr 11 '12 at 16:25
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@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
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Which inequalities did you add to get $\sum\limits_{k=3}^n\frac2{\sqrt k}\le 4\sqrt n +2$ ? I see the telescoping sum $\sum\limits_{k=3}^n\sqrt{k+1}-\sqrt k,$ but how did you get $4\sqrt n+2$? – PinkyWay Nov 29 '20 at 08:54
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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) $\lim\limits_{n\rightarrow\infty} \root n\of n=1$
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$\ \ \ \ \ \ $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.
David Mitra
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