Prove that ${\sqrt {n} }^{\sqrt {n+1}} > {\sqrt {n+1}}^{\sqrt {n}}$.

Question

Which is greater? $\sqrt{n}^{\sqrt{n+1}}$ or $\sqrt{n+1}^\sqrt{n}$

I know that $\sqrt{n}^{\sqrt{n+1}}$ is greater but I tried using induction and I couldn't figure it out. Thanks for the help.

Answer 1

Note that $\frac{\mathrm{d}}{\mathrm{d}x}\frac1x\log(x)=\frac1{x^2}(1-\log(x))<0$ for $x\gt e$. Therefore, $x^{1/x}$ is a decreasing function for $x\gt e$. Thus, for $\sqrt{n}\ge e$, i.e. $n\gt e^2$, $$ \sqrt{n+1}^{1/\sqrt{n+1}}\lt\sqrt{n}^{1/\sqrt{n}}\\ \Updownarrow\\ \sqrt{n+1}^{\sqrt{n}}\lt\sqrt{n}^{\sqrt{n+1}} $$

Answer 2

It actually depends on what $n$ is.

Consider a fixed $a$, and let $f(x)=x^a$ and $g(x)=a^x$. At $x=a$, both functions have equal output. Let's examine their derivatives at $x=a$: $$f'(a) = a^a$$ $$g'(a) = a^a\ln(a)$$ Generally $g$ has the larger derivative at $x=a$, as long as $a>e$. Again as long as $a>e$, after the point at which $f$ and $g$ are equal, $g$ has higher output values.

So let's take $a=\sqrt{n}$ and assume for a bit that $\sqrt{n}>e$. Since $\sqrt{n+1}$ is greater than $a$, we will have that $g(\sqrt{n+1})>f(\sqrt{n+1})$, or rather: $$\sqrt{n}^{\sqrt{n+1}}>\sqrt{n+1}^{\sqrt{n}}$$

For values of $n$ smaller than $e^2$, you could just examine those $n$ values individually. For $n=7$, this direction is still true, but the inequality reverses for $n=0\ldots6$.

Answer 3

This is just another version of "$n^{1/n}$ is decreasing for $n > e$" that pops up here every so often. I and others have provided a number of proofs of this.

Answer 4

Hint: take logarithms of both sides and approximate for sufficiently large $n$.

$$\log{\sqrt{n}^{\sqrt{n+1}}} = \frac{1}{2} \sqrt{n+1} \log{n}$$

For sufficiently large $n$, this is approximately

$$-\frac{1}{2}\sqrt{n} \log \left(\frac{1}{n}\right)-\frac{1}{4} \sqrt{\frac{1}{n}} \log \left(\frac{1}{n}\right)+\frac{1}{16} \left(\frac{1}{n}\right)^{3/2} \log \left(\frac{1}{n}\right)+O\left(\left(\frac{1}{n}\right)^{5/2}\right)$$

$$\log{\sqrt{n+1}^{\sqrt{n}}} = \frac{1}{2} \sqrt{n} \log{(n+1)}$$

For sufficiently large $n$, this is approximately

$$-\frac{1}{2}\sqrt{n} \log \left(\frac{1}{n}\right)+\frac{1}{2}\sqrt{\frac{1}{n}}-\frac{1}{4}\left(\frac{1}{n}\right)^{3/2}+O\left(\left(\frac{1}{n}\right)^{5/2}\right)$$

The latter expression is greater than the former when $n > e^2$. Because the log is monotonic, then

$$\sqrt{n}^{\sqrt{n+1}} > \sqrt{n+1}^{\sqrt{n}}$$

when $n > e^2$.

Answer 5

Here's a general result that might be of some use. If we have two functions, $s(n)\mathrm{\ and\ } b(n)$ ("small" and "big") where $c < s(n) < b(n)$ for some constant $c\ge e$, (as usual, $e$ is the base for the natural log) then $$ b(n)^{s(n)} < s(n)^{b(n)} $$ In other words, roughly speaking, "small-to-the-big beats big-to-the small." In what follows, you might find it helpful to know that $f(n) = n / \log n$ is a strictly increasing function for $n \ge e$.

If $f$ is strictly increasing, we have that if $A < B$ then $f(A) < f(B)$, so since $s(n) < b(n)$ we'll have $$ \frac{s(n)}{\log s(n)} < \frac{b(n)}{\log b(n)} $$ Multiplying both sides of the inequality by the denominators yields $$ s(n)\log b(n) < b(n) \log s(n) $$ and using the power property of logs gives $$ \log(b(n)^{s(n)}) < \log(s(n)^{b(n)}) $$ Since the log function is also strictly increasing we may conclude $$ b(n)^{s(n)} < s(n)^{b(n)} $$ as required. As an example, we can conclude that $n^{\log n} < (\log n)^n$ and, for your question, $\sqrt{n+1}^{\sqrt n} < {\sqrt n}^{\sqrt{n+1}}$ (for all $n$ sufficiently large).