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.

Is it possible to compute it without Taylor series? $$\lim_{n\to\infty}\frac{n}{\ln \ln n}\cdot \left(\sqrt[n]{1+\frac{1}{2}+\cdots+\frac{1}{n}}-1\right) $$

Maybe you try your luck without computational software.

share|improve this question
2  
I do not know, if I were enlightened maybe then I could come up with some answer. –  user67878 Mar 28 '13 at 18:44
    
Here is another problem of mine with no progress: math.stackexchange.com/questions/335089/…. Some ideas? (Just in case) –  Enlightened by God Mar 28 '13 at 20:14
    
What answer do you get for the above limit with the help of computational software? I made some calculations and it seems that the limit could be $1$ but I did not check the result more accurately to state that it should be $1$. –  user67878 Mar 28 '13 at 20:19
    
@Thus As far as I know for some limits Wolfram|Alpha (Pro) offers a complete solution. Anyway, I cannot access it right now, and I do not have Wolfram|Alpha Pro version. –  Enlightened by God Mar 28 '13 at 20:25
    
In fact, I showed only that the limit is $\geq1$, if it exists. –  user67878 Mar 28 '13 at 20:26

7 Answers 7

We know the famous result: $$H_n=\sum_{k=1}^n\frac{1}{k}=\log n+\gamma+o(1)$$ so we find $$\sqrt[n]{1+\frac{1}{2}+\cdots+\frac{1}{n}}-1=\exp(\frac{1}{n}\log\left(\log(n)+\gamma+o(1)\right))-1\sim_\infty\frac{\log\log n}{n}$$ hence $$\lim_{n\to\infty}\frac{n}{\ln \ln n}\cdot \left(\sqrt[n]{1+\frac{1}{2}+\cdots+\frac{1}{n}}-1\right) =\lim_{n\to\infty}\frac{n}{\ln \ln n}\frac{\log\log n}{n}=1.$$

share|improve this answer
    
Thanks, I correct it. –  Sami Ben Romdhane Mar 29 '13 at 2:52
    
Just a nitpick: wouldn't it be better to use either one of $\log$ or $\ln$? –  Pedro Tamaroff Mar 31 '13 at 18:09

$ 1 + \frac 1 2 + ... + \frac 1 n = \log n + O(1)$. So $$ \left( 1 + \frac 1 2 + ... + \frac 1 n \right)^{\frac 1 n} = \exp \left( \frac 1 n \log(\log n + O(1)) \right) = 1 + \frac 1 n \log(\log n) + o \left( \frac {\log(\log n)} n \right). $$ So the answer to your problem is $1$ (but I used Taylor series...)

share|improve this answer

Here is an idiosyncratic proof:

First, applying the squeezing lemma to the inequality $ 1 \leq (H_{n})^{1/n} \leq n^{1/n} $, we know that

$$\lim_{n\to\infty} (H_{n})^{1/n} = 1. \tag{1} $$

Next, as tetori pointed out, we have

$$ \log n \leq H_n \leq 1 + \log n. \tag{2} $$

From this inequality, we have

$$ \log \log n \leq \log H_n \leq \log(1 + \log n) = \log\log n + \log \left( 1 + \frac{1}{\log n} \right) $$

In particular, dividing each sides by $\log \log n$ and taking $n \to \infty$ gives

$$\lim_{n\to\infty} \frac{\log H_n}{\log \log n} = 1. \tag{3} $$

Now it is plain to observe that

\begin{align*} \frac{n}{\log\log n} \{ (H_n)^{1/n} - 1 \} &= \frac{1}{\log\log n} \int_{1}^{H_n} \frac{x^{1/n}}{x} \, dx. \end{align*}

Since

\begin{align*} \log H_n &= \int_{1}^{H_n} \frac{1}{x} \, dx \leq \int_{1}^{H_n} \frac{x^{1/n}}{x} \, dx \leq \int_{1}^{H_n} \frac{(H_n)^{1/n}}{x} \, dx = (H_n)^{1/n} \log H_n, \end{align*}

we have

$$ \frac{\log H_n}{\log\log n} \leq \frac{n}{\log\log n} \{ (H_n)^{1/n} - 1 \} \leq (H_n)^{1/n} \frac{\log H_n}{\log\log n} .$$

Therefore, taking $n\to\infty$ we obtain the desired limit by both $(1)$ and $(3)$.

share|improve this answer
    
@Aryabhata, Thank you. I wanted to show that circumventing the shortcut may require excessive efforts. –  sos440 Mar 29 '13 at 2:24

Doing this any way but Taylor series is in my mind, insane. But still, you can use L'Hopital. The trick is to express

$$[\log{n}]^{1/n} = \exp{\left ( \frac{\log{\log{n}}}{n}\right)}$$

So what is desired is

$$\lim_{n \rightarrow \infty} \frac{n \left [ \exp{\left ( \frac{\log{\log{n}}}{n}\right)} -1 \right ]}{\log{\log{n}}}$$

Using Taylor series, you can eyeball this and see that the limit is $1$. L'Hopital is the next best thing, however, but it does get messy. You have to take derivatives of numerator and denominator separately, as we have a $0/0$ situation. After some messy algebra, we get the equivalent limit:

$$\lim_{n \rightarrow \infty} \left [ \left ( \exp{\left ( \frac{\log{\log{n}}}{n}\right)} -1 \right ) n \log{n} - \exp{\left ( \frac{\log{\log{n}}}{n}\right)} \log{n}\,\log{\log{n}} \right ] + \exp{\left ( \frac{\log{\log{n}}}{n}\right)}$$

Again, you can see that the term in the brackets goes to zero in this limit if you just examine the first term in the series. But, if you are holding fast to ignorance of the behavior of $\exp{z}-1 \sim z$ near the origin, then you could rearrange the term in the brackets so that you get several complicated terms of the for $0/0$. At this point, however, I will depart you in the knowledge that the series is just way, way, way easier.

share|improve this answer
1  
$\lim_{n \rightarrow \infty} \frac{n \left [ \exp{\left ( \frac{\log{\log{n}}}{n}\right)} -1 \right ]}{\log{\log{n}}}=1$ because $\lim_{x\to0}\frac{e^x -1}{x}=1$ –  Enlightened by God Mar 28 '13 at 19:08
    
Thank you, but this isn't exactly what I need. –  Enlightened by God Mar 28 '13 at 19:12
1  
@EnlightenedbyGod: Yes. And I am sorry that this is not what you need. What I hoped to demonstrate, though, is that there are some problems, like this one, where not using that limit is self-defeating. What was it you are after, exactly? You should note that there are two other answers beside this one, neither of which avoid some form of a series expression. –  Ron Gordon Mar 28 '13 at 19:16
    
I'm looking for a pure proof, without any spot of series expansion. Beside that, your proof is useful for those that are interested in finding a solution. –  Enlightened by God Mar 28 '13 at 19:25

At Fisrt, following inequality holds: $$\ln n \le H_n \le \ln n+1$$ Where $H_n$ is Harmonic number. So, we consider following limits: $$\lim_{n\to\infty} \frac{n}{\ln\ln n} (\sqrt[n]{\ln n} -1) \quad\text{and}\quad \lim_{n\to\infty} \frac{n}{\ln\ln n} (\sqrt[n]{1+\ln n} -1) $$

We apply L'Hôpital's rule to calculate this limit. Before applying L'Hôpital's rule, we check this limit satisfies conditions of L'Hôpital's rule.

At First, we prove that $$\lim_{x\to\infty} \frac{\ln\ln x}{x}=0 \quad\text{and}\quad \lim_{x\to\infty} \sqrt[x]{\ln x} =1.$$ To prove these formulas is easy, and $$\lim_{x\to\infty} \frac{(\sqrt[x]{\ln x} -1)'}{(\ln\ln x/x)'}= \lim_{x\to\infty} \frac{\sqrt[x]{\ln x}\cdot (\ln\ln x/x)' }{(\ln\ln x/x)'}=1$$

So By L'Hôpital's rule we get $$\lim_{x\to\infty} \frac{\sqrt[x]{\ln x} -1}{\ln\ln x/x}=\lim_{x\to\infty} \frac{(\sqrt[x]{\ln x} -1)'}{(\ln\ln x/x)'}= 1$$

And you can show that $$\lim_{n\to\infty} \frac{n}{\ln\ln n} (\sqrt[n]{1+\ln n} -1)=1.$$

share|improve this answer
    
Wow, I didn't know that you are also in here! Nice to meet you. And nice vocaloid(?) portrait, too. :) –  sos440 Mar 29 '13 at 2:06

A limit I use a bit, which is the inverse of $\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x$: $$ \lim_{n\to\infty}n\left(x^{1/n}-1\right)=\log(x)\tag{1} $$ Therefore, for any constant $\alpha$, substituting $n\mapsto\frac{n}{\log(\alpha+\log(n))}$, we get $$ \lim_{n\to\infty}\frac{n}{\log(\alpha+\log(n))}\left(x^{\log(\alpha+\log(n))/n}-1\right)=\log(x)\tag{2} $$ Plugging $x=e$ into $(2)$ yields $$ \lim_{n\to\infty}\frac{n}{\log(\alpha+\log(n))}\left((\alpha+\log(n))^{1/n}-1\right)=1\tag{3} $$ Since $1+\frac12+\frac13+\dots+\frac1n\to\gamma+\log(n)$ and $\lim\limits_{n\to\infty}\frac{\log(\alpha+\log(n))}{\log(\log(n))}=1$, $(3)$ gives $$ \lim_{n\to\infty}\frac{n}{\log(\log(n))}\left(\left(1+\frac12+\frac13+\dots+\frac1n\right)^{1/n}-1\right)=1\tag{5} $$

share|improve this answer

Without Taylor series.

First we show that, for $0 \le x \le 1$, we have that

$$x \le e^x -1 \le x + ex^2$$

This can easily be shown by taking derivatives.

For instance, for the right side, let $f(x) = e^x - 1 - x - ex^2$.

We have that $f(0) = 0$.

Now $f'(x) = e^x - 1 -2ex$. Taking derivatives again, we can show that

$f'(x) \le 0$ for $0 \le x \le 1$ (basically $f'(0) = 0$ and $f''(x) \le 0$ for $0 \le x \le 1$)

Thus $f(x) \le 0$ for $0 \le x \le 1$.

Now just use any elementary proof that $\log (n-1) \le H_n \le \log n+ 1$ (say using integrals).

Let $$ S_n = H^{1/n} -1 = e^{\frac{\log H_n}{n}} -1$$

Now using the above inequality (for sufficiently large $n$), we get that

$$ \frac{\log H_n}{n} \le S_n \le \frac{\log H_n}{n} + \frac{e\log^2 H_n}{n^2}$$

Now by mean value theorem, for any $c > 0$ we have that $\log \log (n+c) - \log \log n \le \frac{c}{n \log n}$ and thus $\lim_{n \to \infty} \frac{\log \log (n+c)}{\log \log n} = 1$.

Now it is easy to see that your sequence has the limit $1$.

share|improve this answer
    
Good to see you back. –  Pedro Tamaroff Mar 29 '13 at 4:03
    
@PeterTamaroff: Thanks! –  Aryabhata Mar 29 '13 at 4:04

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.