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I would need a proof that $n \left(1-p^{\frac{1}{n}}\right)$ is increasing in $n \in \mathbf{N}$ for any $p \in (0,1)$.

Context

I am working on a larger question and this is the last missing piece. But with this I'm a bit out of ideas (I tried the difference between $n+1$ and $n$ and also derivative wrt $n$, but neither gave anything that seemed useful). I know how to find limit of this with L'Hospital's rule (it is $-\log p$). But I it does not seem to help to prove the monotonicity.

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    $\begingroup$ What did you get so far, where are you struggling? $\endgroup$ – konewka Oct 16 '14 at 9:58
  • $\begingroup$ I am working on a larger question and this is the last missing piece. But with this I'm a bit out of ideas (I tried the difference between $n+1$ and $n$ and also derivative wrt $n$, but neither gave anything that seemed useful). $\endgroup$ – TomH Oct 16 '14 at 10:08
  • $\begingroup$ I know how to find limit of this with L'Hospital's rule (it is $-\log p$). But I it does not seem to help to prove the monotonicity. $\endgroup$ – TomH Oct 16 '14 at 10:10
  • $\begingroup$ Alright, just to get your question clear, you want to show that $f:\mathbb{N}\rightarrow \mathbb{R}, f(n)=n\left(1-p^{\frac{1}{n}}\right)$ is increasing? $\endgroup$ – konewka Oct 16 '14 at 10:14
  • $\begingroup$ You could use $d/dn$ $\endgroup$ – Alexandre Halm Oct 16 '14 at 10:14
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One could also use elementary estimates.

Fix $n$, and set $p = x^{n(n+1)}$. Now the proving the statement is equivalent to $$ n(1-x^{n+1}) \leq (n+1)(1-x^{n}) $$ which can be written as $$ nx^{n}(1-x) \leq (1-x^{n}). $$ Since $x \in (0, 1)$, we may divide by $1-x$ to get $$ nx^n \leq 1+x+\ldots + x^{n-1}. $$ But since $x^{n} \leq x^{k}$, for each $k \in \{0, 1, \ldots, n - 1\}$, the statement follows. (And the inequality is even strict.)

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  • $\begingroup$ Very nice solution! $\endgroup$ – TomH Oct 16 '14 at 12:48
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Take the derivative to $n$:

$$\frac{d}{dn}n(1-p^{\frac{1}{n}})=1-p^{\frac{1}{n}}+\frac{np^{\frac{1}{n}}\log p}{n^2}=1+\left(\frac{1}{n}\log p-1\right)p^{\frac{1}{n}}$$

This will have to be greater than $0$ for every $n\in\mathbb{N},p\in(0,1)$, so $$1+\left(\frac{1}{n}\log p-1\right)p^{\frac{1}{n}}>0\iff\left(\frac{1}{n}\log p-1\right)p^{\frac{1}{n}}>-1$$ and this holds for all $n$ and for all $p>0$, so the function $n(1-p^{\frac{1}{n}})$ is increasing in $n$

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  • $\begingroup$ Thanks, but I think the proof is still incomplete. I do not see the last step: why does the inequality hold? $\log p < 0$, so the first term in the parenthesis is less than $-1$ and $p^{\frac{1}{n}} \in (0,1)$, so it does not look obvious to me. $\endgroup$ – TomH Oct 16 '14 at 10:49
  • $\begingroup$ Same problem I ran into, as I reread my answer. I've been trying to solve it too, but no breakthrough yet. WolframAlpha says it's true, but I'm not yet able to verify, hang on $\endgroup$ – konewka Oct 16 '14 at 10:56
  • $\begingroup$ I've done it, you can define a function replacing 1/n by u , with the condition u>0 . You derive h(u) ( u <- 1/x in f' ) and you can conclude $\endgroup$ – mvggz Oct 16 '14 at 12:05
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0 < p < 1 : let f : x-> x*(1 - $p^\frac{1}{x}$)

$f'(x) = (1-p^\frac{1}{x}) +\frac{1}{x}*ln(p)*p^\frac{1}{x} = (1-p^\frac{1}{x}) -\frac{1}{x}*ln(\frac{1}{p})*p^\frac{1}{x} $ = $1 - [1+\frac{1}{x}*ln(\frac{1}{p})]*p^\frac{1}{x}$

Let $ h : u-> 1 - [1 + u*ln(\frac{1}{p})]*p^u = f'(\frac{1}{u}) $

$ h'(u) = u*p^u*[ln(p)]^2 >0 $, when u>0

h increases, h(0) = 0, so for u >0, h(u)>0 => $f'(x)$ = $h(\frac{1}{x}) > 0 $ So f increases relatively to x,

n+1 > n => f(n+1) > f(n) => Your sequence increases

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    $\begingroup$ That's a different formula, you should have $f(x)=x(1-p^{\frac{1}{x}})$ $\endgroup$ – konewka Oct 16 '14 at 10:23
  • $\begingroup$ My bad, you're right.. $\endgroup$ – mvggz Oct 16 '14 at 10:26
  • $\begingroup$ I've changed my answer, I think it's good now $\endgroup$ – mvggz Oct 16 '14 at 12:06
  • $\begingroup$ Yeah, that's more like it, hope the OP agrees as well $\endgroup$ – konewka Oct 16 '14 at 12:10
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    $\begingroup$ Thx, I'm sorry I wrote the thing on a rush, I'll be more careful. $\endgroup$ – mvggz Oct 16 '14 at 12:12

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