I have seen this problem in a book .

But I don't know what should be the solution . Question is

There is a sequence defined by $$\sqrt[1]{1},\sqrt[2]{2},\sqrt[3]{3},\sqrt[4]{4},\cdots,\sqrt[n-1]{n-1},\sqrt[n]{n}$$

We have to find the largest number in this sequence.

What I did is as usual what I do when comparing two irrational numbers

I compared $\sqrt[1]{1} $ & $\sqrt[2]{2}$ and found that $\sqrt[2]{2}$ is greater.

Then I compared $\sqrt[3]{3} $ &$\sqrt[2]{2}$ and found $\sqrt[3]{3}$ is greater.

Then I compared $\sqrt[3]{3} $& $\sqrt[4]{4}$ and found $\sqrt[3]{3}$ is greater.

And then I compared $\sqrt[5]{5}$&$\sqrt[4]{4}$ and found $\sqrt[4]{4}$ is greater.

And finally I compared $\sqrt[5]{5}$&$\sqrt[6]{6}$ and found $\sqrt[5]{5}$ is greater.

I cannot guarantee will it work up to infinity or not and found $\sqrt[3]{3}$ is greatest term.

And now I have to prove my result.

More generally I have to prove that $\sqrt[n]{n}\gt \sqrt[n+1]{n+1}$ for all $n\in\mathbb N$ and $n\gt 2$

I applied induction to prove it but didn't got the solution.

Please help me in proving this .

Or please tell me an alternative way to tell which number is largest in the sequence.


marked as duplicate by Martin Sleziak, kingW3, Namaste, Lord Shark the Unknown, AugSB Oct 4 '17 at 19:15

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    $\begingroup$ The greatest is in fact $\sqrt[e]{e}$. Since $e\not\in\mathbb{N}$, we are left with the closest integer, which happens to be $3$. $\endgroup$ – barak manos Dec 9 '16 at 16:34
  • $\begingroup$ Ohh,, nice... but may I have any hint how to continue further in my way because I am not too much familiar with exponential functions $\endgroup$ – Atul Mishra Dec 9 '16 at 16:36

Notice that your inequality is equivalent to $$ n^{n+1} > (n+1)^n = \sum_{k=0}^{n} {n \choose k} n^{n-k}. $$ Now, the sum on the right has $n+1$ terms. A simple upper bound for a generic term is $$ {n \choose k} n^{n-k} \le \frac{n^k}{k!} n^{n-k} = \frac{n^n}{k!}.$$ Thus, we have $$ (n+1)^n \le n^n(1 + 1 + \frac1{2!} + \ldots + \frac1{n!}) \le n^n \cdot e. $$ Using that for $n \geq 3$ we have also $n > e$, we get $(n+1)^n < n^n \cdot n = n^{n+1}.$

  • $\begingroup$ Interesting switch over and nice choice of bounding - the $\frac {\ln x}{x} $ I used was familiar to me as the reverse switch from investigating the relationship between $a^b$ and $b^a$. $\endgroup$ – Joffan Dec 9 '16 at 17:19

$\ln x$ is a monotonic increasing function on the valid range, so the maximum in $f(x) = x^{1/x}$ will be at the same value of $x$ as $\ln(f(x)) = \frac{\ln x}x$.

If you can differentiate, you'll find that $\frac{d}{dx}\frac{\ln x}{x} = \frac{1-\ln x}{x^2}$ has its one-and-only zero at $e$, which is a maximum in $f(x)$. Then in the integers this means that either $2$ or $3$ is the maximum value, which you have already established is $3$.


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