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Prove without calculus that the sequence $$L_{n}=\sqrt[n+1] {(n+1)!} - \sqrt[n] {n!}, \space n\in \mathbb N$$ is strictly decreasing.

Martin Sleziak
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user 1591719
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    In other words one needs to prove that $L_n-L_{n-1}=\sqrt[n+1] {(n+1)!} - \sqrt[n] {n!}-(\sqrt[n] {(n!} - \sqrt[n-1] {(n-1)!})=\sqrt[n+1] {(n+1)!}-2\sqrt[n] {n!}+\sqrt[n-1] {(n-1)!}<0$ – Adi Dani Nov 15 '12 at 22:05
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    Or you could try to prove that $\left(\sqrt[n+1] {(n+1)!} - \sqrt[n] {n!}\right) \big/ \left(\sqrt[n] {n!} - \sqrt[n-1] {(n-1)!}\right) < 1$ for all $n$. – Hans Engler Nov 16 '12 at 01:33
  • Would it help to look at the second derivative of [$\Gamma(n+1)^{1/n}$](http://tinyurl.com/b4o8hmt)? It's negative and goes to $0$ in the limit... (your site-self-promotion is working ;-) – draks ... Jan 14 '13 at 11:52
  • @draks... yeah, I'm a creative person as regards your last remark :-). However, for the problem I need an approach without calculus. – user 1591719 Jan 14 '13 at 11:56
  • sorry, I just starred at the equation without reading the prerequisites... – draks ... Jan 14 '13 at 12:36
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    $\sqrt[n] {n!}$ is the geometric mean of 1..n. – Mark Hurd Jan 23 '13 at 16:05
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    @Mark Hurd AM-GM inequality would be too strong to estimate $\sqrt[n]{n!}$. As well I tried HM, you'll have to deal with the harmonic series, which is rather complicated. – Michael Li Jan 26 '13 at 11:13
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    @Jonas: Perhaps it would be beneficial to firm up what is meant by "without calculus" in the question statement. – cardinal Feb 03 '13 at 20:54
  • @cardinal: I do not have any suggestions in that regard. – Jonas Meyer Feb 04 '13 at 02:57
  • @Jonas: Ok, fair enough. One thing that (immediately) comes to mind: Are we admitting any use at all of the exponential function? – cardinal Feb 04 '13 at 03:05
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    @cardinal: As far as I am concerned, yes. – Jonas Meyer Feb 04 '13 at 03:41
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    Can we use the fact that $\sqrt[n] {n!}$ is a stictly increasing function? – Koba Feb 06 '13 at 17:11
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    @Jonas: For what it's worth, the sequence $L_n$ is called [Lalescu's sequence](http://en.wikipedia.org/wiki/Traian_Lalescu) and is known to converge to $1/e$. – Mike Spivey Feb 06 '13 at 20:01
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    @Mike: Thanks, it is interesting at least to learn that it has a name! Here's a related question I just remembered: http://math.stackexchange.com/q/161682/ – Jonas Meyer Feb 06 '13 at 21:08
  • One of the things Google finds for Lalescu's sequence is [this answer](http://math.stackexchange.com/a/178798/) by Chris's sister. – Jonas Meyer Feb 06 '13 at 21:35
  • It's also interesting that I see answers from new (or less used) MSE accounts. – user 1591719 Feb 06 '13 at 22:20
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    I couldn't get it to work, but this seems like a good candidate for induction. – Adam Saltz Feb 07 '13 at 05:17
  • Even though it didn't work for me, maybe this could help: $$e\left(\frac ne\right)^n \leq n! \leq e\left(\frac{n+1}e\right)^{n+1}$$ – Thomas Feb 08 '13 at 09:49
  • @macydanim: can this inequality be proved without calculus? – user 1591719 Feb 08 '13 at 09:52
  • Hmm the only proof i found does use integration. http://en.wikipedia.org/wiki/Factorial#Rate_of_growth_and_approximations_for_large_n – Thomas Feb 08 '13 at 10:31
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    I think your question is meaning less: I doubt you can construct (I mean really *construct*) any totally ordered nth-root-closed field without calculus (topology, analysis or whatever you call things based on Cauchy sequences). – Lierre Feb 12 '13 at 22:50

2 Answers2

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Let $\ell_n = \left(n!\right)^{1/n}$. Clearly for all $n \in \mathbb{N}$, $\ell_n > 0$. The question is equivalent to showing that $$\frac{\ell_{n+2}}{\ell_{n+1}} + \frac{\ell_n}{\ell_{n+1}} < 2 \tag{1}$$ Let $$ x_n = \log \frac{\ell_{n+1}}{\ell_n} = \frac{1}{n+1} \left( \log(n+1) - \frac{1}{n} \sum_{k=1}^n \log(k) \right) $$ The inequality $(1)$ now reads: $$ 2 > \exp(x_{n+1}) + \exp(-x_n) = 2 \exp\left(\frac{x_{n+1}-x_n}{2}\right) \cosh \left(\frac{x_{n+1}+x_n}{2}\right) \tag{2} $$ We can rewrite $x_n$ a little: $$ x_n = \frac{1}{n+1} \left( \log\left(\frac{n+1}{n}\right) - \underbrace{\frac{1}{n} \sum_{k=1}^n \log\left(\frac{k}{n}\right)}_{\text{denote this as } s_n} \right) $$

Note that, with some straightforward algebra $$ \frac{x_{n+1}-x_n}{2} = \frac{1}{2(n+2)} \log\left(1+\frac{1}{n+1}\right) - \frac{1}{(n+1)(n+2)} \left( \log\left(1+\frac{1}{n} \right) - s_n \right) \tag{3} $$ $$ \frac{x_{n+1}+x_n}{2} = \frac{1}{2(n+2)} \log\left(1+\frac{2}{n}\right)+ \frac{1}{2(n+2)} \log\left(1+\frac{1}{n}\right) - \frac{1}{n+2} s_n \tag{4} $$

Bounding $s_n$

Using summation by parts: $$ \sum_{k=1}^n \left(a_{k+1}-a_k\right) b_k = a_{n+1} b_n - a_1 b_1 -\sum_{k=1}^{n-1} a_{k+1} \left(b_{k+1} - b_k \right) $$ with $a_k = \frac{k}{n}$ and $b_k = \log \frac{k}{n}$, we find $$ \begin{eqnarray} s_n &=& 0 - \frac{\log n^{-1}}{n} - \sum_{k=1}^{n-1} \frac{k+1}{n} \log\left(1+\frac{1}{k}\right) \\ &=& -\frac{n-1}{n} + \frac{1}{2} \frac{\log(n)}{n} - \frac{1}{n} \sum_{k=1}^{n-1} \left( \left(k + \frac{1}{2} \right) \log\left(1+\frac{1}{k}\right) - 1 \right) \end{eqnarray} $$ Using elementary integral $\int_0^1 \frac{\mathrm{d}x}{k+x} = \log\left(1+\frac{1}{k}\right)$ we find $$ \left(k + \frac{1}{2} \right) \log\left(1+\frac{1}{k}\right) - 1 = \int_0^1 \left(\frac{k+\frac{1}{2}}{k+x}-1\right) \mathrm{d}x = \int_0^1 \frac{1-2x}{2(k+x)} \mathrm{d}x $$ changing variables $x \to 1-x$ and averaging with the original: $$\begin{eqnarray} \left(k + \frac{1}{2} \right) \log\left(1+\frac{1}{k}\right) - 1 &=& \int_0^1 \frac{ \left(x-\frac{1}{2}\right)^2}{(k+x)(k+1-x)} \mathrm{d}x \\ &=& \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{ u^2}{\left(k+\frac{1}{2}\right)^2 - u^2 } \mathrm{d} u \end{eqnarray} $$ Since $$ \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{ u^2}{\left(k+\frac{1}{2}\right)^2 } \mathrm{d} u < \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{ u^2}{\left(k+\frac{1}{2}\right)^2 - u^2 } \mathrm{d} u < \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{ u^2}{\left(k+\frac{1}{2}\right)^2 - \frac{1}{4} } \mathrm{d} u $$ We have $$ \frac{1}{12} \frac{1}{\left(k+\frac{1}{2}\right)^2} < \left(k + \frac{1}{2} \right) \log\left(1+\frac{1}{k}\right) - 1 < \frac{1}{12 k(k+1)} = \frac{1}{12 k} - \frac{1}{12 (k+1)} $$ Since $$\frac{1}{12} \frac{1}{\left(k+\frac{1}{2}\right)^2} > \frac{1}{12} \frac{1}{\left(k+\frac{1}{2}\right) \left(k+\frac{3}{2}\right)} = \frac{1}{12} \frac{1}{k+\frac{1}{2}} - \frac{1}{12} \frac{1}{k+\frac{3}{2}} $$

We thus establish that $$ \sum_{k=1}^{n-1} \left( \left(k + \frac{1}{2} \right) \log\left(1+\frac{1}{k}\right) - 1 \right) < \sum_{k=1}^{n-1} \left( \frac{1}{12 k} - \frac{1}{12 (k+1)} \right) = \frac{1}{12} - \frac{1}{12 n} < \frac{1}{12} $$ and $$ \sum_{k=1}^{n-1} \left( \left(k + \frac{1}{2} \right) \log\left(1+\frac{1}{k}\right) - 1 \right) > \sum_{k=1}^{n-1} \left( \frac{1}{12 \left(k+\frac{1}{2}\right)} -\frac{1}{12 \left(k+\frac{3}{2}\right)} \right) = \frac{1}{18} - \frac{1}{6 (2n+1)} = \frac{1}{9} \frac{n-1}{2n+1} $$ The argument above suggests that $ \sum_{k=1}^{n-1} \left( \left(k + \frac{1}{2} \right) \log\left(1+\frac{1}{k}\right) - 1 \right)$ converges to a number $c$ such that $\frac{1}{18} < c < \frac{1}{12}$. Thus $$ -\frac{n-1}{n} + \frac{\log(n)}{2n} - \frac{1}{12 n} < s_n < -\frac{n-1}{n} + \frac{\log(n)}{2n} - \frac{1}{9} \frac{n-1}{n (2n+1)} \tag{5} $$ Implying that $s_n$ converges to $-1$ and that, for large $n$ $$ s_n = -1 + \frac{\log(n)}{2n} + \mathcal{O}\left(n^{-1}\right) $$

Using these bounds

We therefore conclude that $\frac{x_{n+1}-x_n}{2} = \mathcal{O}\left(n^{-2}\right)$ and $\frac{x_{n+1}+x_n}{2} = \mathcal{O}\left(n^{-1}\right)$. Since both the mean and difference are arbitrarily small for large enough $n$: $$ \begin{eqnarray} 2 \exp\left(\frac{x_{n+1}-x_n}{2}\right) \cosh \left(\frac{x_{n+1}+x_n}{2}\right) &<& 2 \frac{1}{1-\frac{x_{n+1}-x_n}{2}} \frac{1}{1-\frac{1}{2} \left(\frac{x_{n+1}+x_n}{2}\right)^2} \\ &=& 2 + 2\left(\frac{x_{n+1}-x_n}{2}\right) + \left(\frac{x_{n+1}+x_n}{2}\right)^2 + \mathcal{o}\left(n^{-3}\right) \\ &=& 2 - \frac{1}{2 n^3} + \mathcal{o}\left(n^{-3}\right) \end{eqnarray} $$

Thus, at least for $n$ large enough the sequence $L_n$ is decreasing.

This painstaking exercise just makes one appreciate the power of calculus.

Sasha
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    At first I thought: "man, this proof is invalid because you are using integral" but later I recalled that the most natural definition for logarithm is that using integral. Fortunately the bounty comes from Jonas and not from Chris's sister, otherwise he would reject your worthy proof. Congratulations. – Matemáticos Chibchas Feb 08 '13 at 20:32
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    I have never seen a bounty that large! –  Nov 11 '15 at 05:16
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    @0.5772156649... There were 2 bounties awarded, by different users. – Sasha Nov 11 '15 at 13:36
  • It also seems that this is the answer with (currently) the largest cumulative bounty earned in this site. – The Amplitwist Nov 27 '21 at 12:25
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My proof attempt contained a fatal mistake. It seems to me that correcting it would be as hard as proving the statement from scratch, so I've decided to give up. The following two lemmas remain valid, and although strictly weaker than the desired result, they might still be useful to someone, so I've decided to leave them here. (Thanks to mercio for pointing out the mistake and apologies to everyone for the inconvenience.)

It is convenient to define $a_n=\sqrt[n]{n!}$ for $n\in\mathbb N$. Then, $L_n=a_{n+1}-a_n$ holds for $n\in\mathbb N$. We define a new sequence by $K_n=\frac{L_n}{a_n}$. Note that $K_n=\frac{a_{n+1}}{a_n}-1$.

Lemma 1. The sequence $(a_n)_{n=1}^\infty$ is strictly increasing.

Proof. Note that $a_{n+1}^{n+1}=(n+1)a_n^n$ holds for all $n\in\mathbb N$ by definition of $(a_n)_{n=1}^\infty$. This implies that $(\frac{a_{n+1}}{a_n})^n=\frac{n+1}{a_{n+1}}$ holds for all $n\in\mathbb N$. But $a_{n+1}^{n+1}=(n+1)!<(n+1)^{n+1}$, therefore we have $a_{n+1}<n+1$ or equivalently $\frac{n+1}{a_{n+1}}>1$, proving the claim. $\square$

Lemma 2. The sequence $(K_n)_{n=1}^\infty$ is strictly decreasing.

Proof. Clearly, it suffices to prove that $(K_n+1)_{n=1}^\infty$ is strictly decreasing. By definition of $K_n$, this means that we have to show that $\frac{a_{n+2}}{a_{n+1}}<\frac{a_{n+1}}{a_n}$ holds for all $n\in\mathbb N$. This is clearly equivalent to showing $a_{n+2}a_n<a_{n+1}^2$, which is the same as showing that $a_{n+2}^{n+2}a_n^{n+2}<a_{n+1}^{2n+4}$ holds for all $n$. By definition of $(a_n)_{n=1}^\infty$ this is equivalent to $(n+2)!n!a_n^2<((n+1)!)^2a_{n+1}^2$, so we only have to show that $$\frac{a_n^2}{a_{n+1}^2}<\frac{n+1}{n+2}$$ is true for all $n$. We can easily establish this fact by induction. Clearly, it holds for $n=1$. Suppose now, it holds for some $n\in\mathbb N$. We will show that it must also hold for $n+1$. To do this, note that $$\begin{align}\Biggl(\frac{a_{n+1}^2}{a_{n+2}^2}\Biggr)^{n+2}&=\frac{((n+1)!)^2a_{n+1}^2}{((n+2)!)^2}\\ &=\frac{(n!)^2(n+1)^2a_{n+1}^2}{((n+1)!)^2(n+2)^2}\\ &=\frac{a_n^{2n}(n+1)^2a_{n+1}^2}{(a_{n+1})^{2n+2}(n+2)^2}\\ &=\Biggl(\frac{a_n^2}{a_{n+1}^2}\Biggr)^n\frac{(n+1)^2}{(n+2)^2}\overset{\text{I.H.}}{<}\Biggl(\frac{n+1}{n+2}\Biggr)^{n+2}<\Biggl(\frac{n+2}{n+3}\Biggr)^{n+2}.\end{align}$$ Here "I.H." denotes the place where we used the inductive hypothesis. Taking $(n+2)$-nd roots completes the induction. $\square$

Dejan Govc
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    But, don't you have $L_n = K_n a_n$ from the definition of $K_n$ ? at the end this get switched to $L_n = K_n / a_n$. – mercio Feb 08 '13 at 16:14
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    Good observation mercio, it becomes $L_{n+1}\over L_n$$=$${K_{n+1}a_{n+1}}\over K_na_n$ and $K_{n+1}\over K_n$$<1$ but $a_{n+1}\over a_n$$>1$ so nothing from this could be said about the monotonicity of the $L_n$. – A.P. Feb 08 '13 at 16:28
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    What a pity, I saw the light at the end of the tunnel, but it is true, the last proposition is flawed :-( – Matemáticos Chibchas Feb 08 '13 at 16:31