I have a problem with the calculation of the following limit. \begin{equation} \lim_{n\rightarrow \infty} \frac{1+\sqrt{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n} \end{equation} I do not know where to start! Thank you very much
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Do you know the limit of $n^{1/n}$? Also $n=\sum_{i=1}^n 1$. – Johan Nov 19 '12 at 13:54
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4Use $\sqrt[n]{n}\to1$, see e.g. here, and Cesaro mean. – Martin Sleziak Nov 19 '12 at 14:12
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As $n$ grows large, what you do when you co from case $n$ to case $n+1$ is adding $1$ to the denominator and adding ever so slightly more than $1$ to the numerator. Intuition dictates it tends towards $1$, and once you have a convergence candidate, you're halfway there. – Arthur Nov 19 '12 at 14:17
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See also: Evaluating Limit Question $\lim\limits_{n\to \infty}\ \frac{1+\sqrt[2]{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}=1$? – Martin Sleziak Apr 17 '18 at 07:14
2 Answers
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There are at least two possibilities. In all of them you use that $\lim_{n\to\infty}=\sqrt[n]{n}=1$.
The first one uses the following result: if $a_n$ is a convergent sequence, then $$\lim_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n}=\lim_{n\to\infty}a_n.$$
The second is to use Stolz's criterion.
Cameron Buie
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Julián Aguirre
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I remember that there's some theorem saying that, given 2 positive sequences $(a_n); (b_n)$ if $\lim \frac{a_n}{b_n} = \alpha$, then $\lim \frac{\mathop\sum_{i = 1}^n a_i}{\mathop\sum_{i = 1}^n b_i} = \alpha$. Is this correct, and what's the name of this theorem? – user49685 Nov 19 '12 at 14:27
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@user49685 It is (a version of) Stolz-Cesaro theorem, see e.g. this answer. – Martin Sleziak Nov 19 '12 at 14:34
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Hint: Note that $$\sqrt[n]{n} \rightarrow 1.$$
You'll need to know and use that fact.
amWhy
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