Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to prove $\left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)^n$ is decreasing for $n\ge 2$. This popped up in some question I was working with. I tried computing the derivative and show its negative but its way too messy. Now I am trying to prove the ratio of terms is $\ge1$ that is $$\frac{\left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)^n}{\left(\frac{\sqrt{n+1}-1}{\sqrt{n+1}}\right)^{n+1}}\ge 1 $$ Which can be reduced to $$\frac{(\sqrt{n+1}\sqrt{n}-1)^n}{(\sqrt{n}(\sqrt{n+1}-1))^n}\frac{\sqrt{n+1}}{\sqrt{n+1}-1}$$ How should I continue?

share|improve this question
    
Take logarithm, and approximate the $\ln (1 - n^{-1/2})$ by a series. –  vonbrand Mar 23 at 20:27
1  
@vonbrand Using approximations yields only asymptotic results. –  Did Mar 23 at 20:56
    
@Did, expand in series then, and see which way it goes. –  vonbrand Mar 23 at 21:02
    
@vonbrand The case when all the terms are positive excepted, it is not so easy to determine the sense of variation of a series. Furthermore, in this case, the function which extends naturally the sequence is not everywhere decreasing. –  Did Mar 23 at 21:20
    
Since $sqrt(n) - 1 < sqrt(n)$ for $n >= 2$, can't you bound that with some value $0 < a < 1$ and show that $a^n$ is decreasing? –  Michael Deardeuff Mar 24 at 6:20
add comment

2 Answers 2

up vote 3 down vote accepted

The $n$th term is $x_n=\exp(u(1/\sqrt{n}))$ for $$u(x)=\log(1-x)/x^2.$$The derivative $u'(x)$ has the sign of $$v(z)=2\log(z)+1-z,$$ with $z=1/(1-x)$. Note that $v'(z)=(2-z)/z$ is positive on $z\lt2$ and negative on $z\gt2$ and that $v(3)=2\log(3)-2=2\log(3/\mathrm e)\gt0$ because $3\gt\mathrm e$ hence $v(z)\gt0$ for every $1\lt z\lt3$. This implies that the function $u$ is increasing on $(0,\frac23)$ and that the sequence $(x_n)$ is decreasing on $n\geqslant\left(\frac32\right)^2$, that is, $n\geqslant3$. Computing $x_2$ and $x_3$ completes the proof.

share|improve this answer
add comment

If I can suggest an alternate approach: think about logarithms! The funcion $w\mapsto\log(w)$ is increasing in $w$; so, $\log(a_n)$ is decreasing if and only if $a_n$ is decreasing.

But $$ \log\left[\left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)^n\right]=n\left[\log(\sqrt{n}-1)-\log(\sqrt{n})\right], $$ which may be (much) easier to work with than your original expression.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.