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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.

Cortizol
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Enlightened by God
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    I do not know, if I were enlightened maybe then I could come up with some answer. –  Mar 28 '13 at 18:44
  • Here is another problem of mine with no progress: http://math.stackexchange.com/questions/335089/a-double-inequality-with-binomials. 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 $1$. –  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. –  Mar 28 '13 at 20:26
  • @Thus And you can upvote me to put some reputation in my pocket. – Enlightened by God Mar 28 '13 at 20:30
  • No problem. Maybe I should prompt other users to do the same. –  Mar 28 '13 at 20:32
  • @Thus: Brilliant job! – Enlightened by God Mar 28 '13 at 20:33
  • Thank you, I was born with a gift to upvote the users which are in a need for themselves to be upvoted by me. –  Mar 28 '13 at 20:35
  • You do know that Taylor Series can sometimes be written as inequalities by truncating a couple of terms, and a proof of that is easily done using "elementary" calculus? – Aryabhata Mar 29 '13 at 02:47

8 Answers8

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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 $$\lim_{n\to\infty}\frac{n}{\ln \ln n}\cdot \left(\sqrt[n]{1+\frac{1}{2}+\cdots+\frac{1}{n}}-1\right) =\lim_{n\to\infty}\frac{n}{\ln \ln n}\frac{\log\log n}{n}=1.$$

4

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 $(3)$.

Sangchul Lee
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$ 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...)

roger
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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.

Ron Gordon
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    $\lim_{n \rightarrow \infty} \frac{n \left [ \exp{\left ( \frac{\log{\log{n}}}{n}\right)} -1 \right ]}{\log{\log{n}}}=1$ 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
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    @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
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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 $$\lim_{n\to\infty} \frac{n}{\ln\ln n} (\sqrt[n]{1+\ln n} -1)=1.$$

Hanul Jeon
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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 $1$.

Aryabhata
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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 $$ \lim_{n\to\infty}\frac{n}{\log(\log(n))}\left(\left(1+\frac12+\frac13+\dots+\frac1n\right)^{1/n}-1\right)=1\tag{5} $$

robjohn
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We begin by noticing that the expression under root sign is $n$th Harmonic number, so rewrite the problem $$\lim_{n \to \infty} \frac {n} {\log \log n} \left(\sqrt [n] {H_n} - 1 \right).$$ Observe that $\log n < H_n < \log n + 1$.

Let $x = \log n$, and $x \to \infty$. This means that $$H_n^{1/n} = \exp \left(\frac {\log x} {\exp x}\right) + o \left (\frac {1} {\exp x}\right).$$ The problem becomes $$\lim_{x \to \infty} \exp \left(\frac {\log x} {\exp x}\right) + o \left(\frac {1} {\log x} \right) = \exp 0 + 0 = 1.$$