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

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The discussion centers on proving the inequality \(n^{n+1} > (n+1)^n\) for all natural numbers \(n \geq 3\). Participants share various approaches to the proof, highlighting mathematical reasoning and techniques. Notable contributors, castor28 and kaliprasad, provide correct solutions that are acknowledged in the thread. The conversation emphasizes the importance of understanding exponential growth in the context of the inequality. The thread serves as a platform for mathematical problem-solving and community engagement.
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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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Congratulations to the following members for their correct solution:

1. castor28
2. kaliprasad

Solution from castor28:
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:
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 $n\in \Bbb{N}$.
 
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