Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
Do you know the limit of $n^{1/n}$? Also $n=\sum_{i=1}^n 1$. –  Johan Nov 19 '12 at 13:54
4  
Use $\sqrt[n]{n}\to1$, see e.g. here, and Cesaro mean. –  Martin Sleziak Nov 19 '12 at 14:12
    
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

3 Answers 3

up vote 7 down vote accepted

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.

share|improve this answer
    
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
    
@user49685 It is (a version of) Stolz-Cesaro theorem, see e.g. this answer. –  Martin Sleziak Nov 19 '12 at 14:34

Hint: Note that $$\sqrt[n]{n} \rightarrow 1.$$

You'll need to know and use that fact.

share|improve this answer
    
@DonAntonio I second that. –  Dan Shved Nov 19 '12 at 14:08

Hint: It's easy. 1000^(1/1000) ~ 1. Use the algebra of limits: lim n-> infinity (a_n/b_n) = (lim n->infinity a_n)/(lim n->infinity b_n). The numerator tends to n (plus some constant) and the denominator tends to n. So the limit is 1.

share|improve this answer
    
That is incorrect. $\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = \frac{\lim_{n \rightarrow \infty}a_n}{\lim_{n \rightarrow \infty}b_n}$ if both limits are finite and denominator limit is not 0. Also What does "numerator tend to n" mean when you take limit to infinity? –  Gautam Shenoy Nov 19 '12 at 14:06
    
I mean you get S_n = 1 + (close to 1) + (close to 1) + ... n times which is approximately n. If you want more rigour, I only need to mention that n^(1/n) tends to 1 as n tends to infinity. That way, given epsilon, there exists an n_0 such that n^(1/n) - 1 is less than epsilon for all n >= n_0. Or in other words, S_n gets relatively closer to n as n tends to infinity. When n is 1000 S_n is relatively close to n = 1000. When n is 1,000,000, S_n is relatively even closer to n = 1,000,000 than it was for n = 1000. This is all that matters here. –  Adam Rubinson Nov 19 '12 at 14:30
    
If (a_n) is a sequence with a_n-->1 as n-->infinity then lim(k-->infinity) [Sum(a_1 to a_k)] / k = 1. –  Adam Rubinson Nov 19 '12 at 14:37
    
@AdamRubinson Of course this result is true. The argument in your answer is rather poor, though. –  Cocopuffs Nov 19 '12 at 14:42
    
Sure. For stuff like this I think intuition is more important than rigour. But maybe that's personal taste –  Adam Rubinson Nov 19 '12 at 14:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.