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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?

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Take logarithm, and approximate the $\ln (1 - n^{-1/2})$ by a series. – vonbrand Mar 23 '14 at 20:27
@vonbrand Using approximations yields only asymptotic results. – Did Mar 23 '14 at 20:56
@Did, expand in series then, and see which way it goes. – vonbrand Mar 23 '14 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 '14 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 '14 at 6:20
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.

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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.

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