Prove elementarily that $\sqrt[n+1] {(n+1)!} – \sqrt[n] {n!}$ is strictly decreasing

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

Answer

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.

Attribution
Source : Link , Question Author : user 1591719 , Answer Author : Sasha

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