How can one show that $n^{\frac{n+1}{n}}$ is increasing, possibly using logarithms but not derivatives? I obtain $(n^2+2n)log(n+1)\geq (n^2+2n+1)log (n)$ but cannot proceed further.


A purely algebraic proof

We have to prove that $$\bigl(n+1\bigr)^{\tfrac{n+2}{n+1}}> n^{\tfrac{n+1}{n}}\iff (n+1)^{n(n+2)}>n^{(n+1)^2}.$$ Note that $\;(n+1)^{n(n+2)}=(n+1)^{(n+1)^2-1} $, so the last inequality is equivalent to $$\frac{(n+1)^{(n+1)^2-1}}{n^{(n+1)^2}}>1\iff\Bigl(1+\frac1n\Bigr)^{(n+1)^2}>n+1$$ Let's apply Bernoulli's inequality: $$\Bigl(1+\frac1n\Bigr)^{(n+1)^2}>1+\frac{(n+1)^2}{n}=1+n+2+\frac 1n>3+n.$$

  • $\begingroup$ Well done! (+1) $\endgroup$ – Mark Viola Sep 19 '17 at 22:50
  • $\begingroup$ (+1) I wrote an answer, but when I simplified it, it became very similar to yours. $\endgroup$ – robjohn Sep 23 '17 at 15:49
  • $\begingroup$ @robjohn: Great minds think together! ;o) $\endgroup$ – Bernard Sep 23 '17 at 17:39

Hint : Use the fact that $\ln(n+1) \geq \ln(n) +\frac{1}{n+1}$

Lemma : $\ln(n+1) \geq \ln(n)+\frac{1}{n+1}$

Or we have $\ln(\frac{n+1}{n}) \geq \frac{1}{n+1}$

Exponent-ate both side gives us that $1+\frac{1}{n} =1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\cdots\geq e^{\frac{1}{n+1}} = 1+\frac{1}{n+1}+\frac{1}{2(n+1)^2}+\cdots$

Visually the proof is obvious.

  • $\begingroup$ How do I prove that? $\endgroup$ – HBHSU Sep 19 '17 at 22:33
  • $\begingroup$ you want a complete proof ?! $\endgroup$ – Ahmad Sep 19 '17 at 22:33
  • $\begingroup$ I simply cannot find a simple proof for my question, which seems very simple. $\endgroup$ – HBHSU Sep 19 '17 at 22:35
  • 2
    $\begingroup$ @HBHSU The last inequality is true because the coefficient each term of the geometric series (on the LHS) is greater than the corresponding coefficient of the exponential series (on the RHS); more formally one can write $1+\frac1n-e^{1/(n+1)}$ as $\sum_{i=2}^\infty (1-\frac1{i!})(n+1)^{-i}$ and then note that every term in this sum is manifestly positive. $\endgroup$ – Steven Stadnicki Sep 19 '17 at 23:12
  • 1
    $\begingroup$ then you will have that $(n^2+2n)\ln(n+1) \geq (n^2+2n)(\ln n+\frac{1}{n+1}) \geq (n^2+2n+1) \ln n$ so you get to $\frac{n^2}{n+1}+n^2 \ln(n)+\frac{2 n}{n+1}+2 n \ln(n) \geq n^2 \ln(n)+2 n \ln(n)+\ln(n)$ , now don't tell me this is not clear. $\endgroup$ – Ahmad Sep 19 '17 at 23:12

How about this:

Since $n^{1+1/n}=e^{(1+1/n)\ln{n}}$, and the exponential is a monotonically increasing function, so is $(1+1/n)\ln{n}$. Then verify that the function $g(x)=(1+1/x)\ln x$ is monotonically increasing. One has only to prove that $g'(x)>0$ $\forall x>0$. We have \[ g'(x)= -\frac{1}{x^2}\ln x+\left(1+\frac{1}{x}\right)\frac{1}{x} \] Then, using the inequality $-\ln x\geq 1-x$, valid for $x>0$, we obtain \[ g'(x)\geq (1-x)\frac{1}{x^2}+\left(1+\frac{1}{x}\right)\frac{1}{x}=\frac{2}{x^2}>0 \]


Note (added after posting): I didn't notice the OP's proscription on derivatives. I'm leaving the answer in case anyone wants to see an approach ignoring the proscription.

Let $u=1/n$. Then

$$n^{n+1\over n}=n^{1+{1\over n}}=\left(1\over u\right)^{1+u}={1\over u^{1+u}}$$

It suffices to show that $u^{1+u}$ is an increasing function of $u$, which is equivalent to showing that $f(u)=(1+u)\ln u$ is an increasing function:

$$f'(u)={1+u\over u}+\ln u={1+u\over u}+\int_1^u{dt\over t}\ge{1+u\over u}+\int_1^u{dt\over u}={1+u\over u}+{u-1\over u}=2\gt0$$

Remark: The inequality

$$\int_1^u{dt\over t}\ge\int_1^u{dt\over u}$$

is easy to see for $u\ge1$, since $1/t$ is being replaced by its smallest value, $1/u$. For $0\lt u\lt1$ it's a bit trickier: You're replacing $1/t$ with its largest value, but you're integrating "backwards," producing a negative number.


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