Is \(n^{n+1} > (n+1)^n\) for All \(n \geq 3\)?

[anemone]

Here is this week's POTW:

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Prove that $n^{n+1}>(n+1)^n$ for all $n\ge 3$ and $n\in \Bbb{N}$.

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[anemone]

Congratulations to the following members for their correct solution:

1. castor28
2. kaliprasad

Solution from castor28:

Spoiler

As the relation is obviously satisfied for $n=3$, we will prove the result by showing that the LHS grows faster than the RHS. We will allow $n$ to take any real value greater than or equal to 3. According to usual convention, we will write $x$ instead of $n$, and we will assume that $x\ge3$.

Because $\ln$ is an increasing function, we may take logarithms, and we must prove that:
$$h(x) = (x+1)\ln(x) - x\ln(x+1) > 0\qquad[1]$$
for $x\ge3$.

Differentiating, we get:
$$h'(x) = \ln(x) + \frac{x+1}{x} - \ln(x+1) - \frac{x}{x+1}$$
We note that $h'(x)\to0$ when $x\to\infty$. We now compute the second derivative:
$$\begin{align*}
h''(x) &= \frac1x + \frac1x -\frac{x+1}{x^2} - \frac{1}{x+1} - \frac{1}{x+1} + \frac{x}{(x+1)^2}\\
&= \frac2x -\frac{x+1}{x^2}- \frac{2}{x+1}+ \frac{x}{(x+1)^2}\\
&= -\frac{x^2+x+1}{x^4+2x^3+x^2}\\
&= -\frac{x^2+x+1}{x^2(x+1)^2}
\end{align*}
$$
As the numerator has no real root, this shows that $h''(x)<0$, i.e., $h'(x)$ is a decreasing function. As $h'(x)\to0$ when $x\to\infty$, $h'(x)>0$ and $h(x)$ is an increasing function.

As $h(3) = 4\ln(3)-3\ln(4)\approx 0.235>0$, this implies that $h(x)>0$ for $x\ge3$; this is the relation [1] that we needed to prove.

Alternative Solution:

Spoiler

From the formula $$(1+x)^n=1+\frac{nx}{1!}+\frac{n(n-1)x^2}{2!}+\frac{n(n-1)(n-2)x^3}{3!}+\cdots$$, if we replace $x$ by $$\frac{1}{n}$$ we get

$$\left(1+\dfrac{1}{n}\right)^n\le 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots\le 1+1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{3}+\cdots=3\le n$$

$\therefore n^{n+1}>(n+1)^n$ for $n\ge 3$ where $n\in \Bbb{N}$.