Show without differentiation that $\frac {\ln{n}}{\sqrt{n+1}}$ is decreasing

Question

Show that the function $\displaystyle \frac {\ln{n}}{\sqrt{n+1}}$ is decreasing from some $n_0$

My try: $\displaystyle a_{n+1}=\frac{\ln{(n+1)}}{\sqrt{n+2}}\le \frac{\ln{(n)+\frac{1}{n}}}{\sqrt{n+2}}$

so we want to show that $\ln{n}\cdot(\sqrt{n+1}-\sqrt{n+2})+\frac{\sqrt{n+1}}{n}\le 0$ or equivalently $n\cdot \ln{n} \cdot (\sqrt{\frac{n+2}{n+1}}-1)\ge1$ and I'm stuck here.

Answer 1

First, you can show very easily without calculus that $k!>4^k$ whenever $k\ge18$: we then have $$\eqalign{k! &=(1\times2\times3)\times(4\times5\times\cdots\times15)\times(16\times17\times18)\times19\times\cdots\times k\cr &>(1\times1\times1)\times(4\times4\times\cdots\times4)\times(4^2\times4^2\times4^2)\times4\times\cdots\times4\cr &=4^k\ .\cr}$$ (In fact, this is true for $k\ge9$, but the proof takes a little more work.) Now for any $n$ we have $$\eqalign{\Bigl(1+\frac1n\Bigr)^{2n+3} &=\sum_{k=0}^{2n+3}\binom{2n+3}k\frac1{n^k}\cr &\le\sum_{k=0}^{2n+3}\frac{(3n)^k}{k!}\frac1{n^k}\cr &\le\sum_{k=0}^{17}\frac{3^k}{k!}+\sum_{k=18}^\infty\Bigl(\frac34\Bigr)^k\cr &=\Bigl(\sum_{k=0}^{17}\frac{3^k}{k!}\Bigr)+3\Bigl(\frac34\Bigr)^{17}\ .\cr}$$ This last quantity is a specific, fixed number. So if $n$ is large enough we have $$\eqalign{ n>\Bigl(1+\frac1n\Bigr)^{2n+3} &\Rightarrow n^{2n+4}>(n+1)^{2n+3}\cr &\Rightarrow n^{n+2}>(n+1)^{n+\frac32}\cr &\Rightarrow n^{n+2}>(n+1)^{\sqrt{(n+1)(n+2)}}\cr &\Rightarrow n^{1/\sqrt{n+1}}>(n+1)^{1/\sqrt{n+2}}\cr &\Rightarrow \frac{\log n}{\sqrt{n+1}}>\frac{\log(n+1)}{\sqrt{n+2}}\ .\cr}$$

Answer 2

If you subtract f(n+1)-f(n), then it comes down to $\frac{(n+1)^{\sqrt{n+1}}}{(n)^{\sqrt{n+2}}}\leq 1$ , take logs and it should be easy to finish?

Answer 3

Edit: OK, so it turns out I just proved the given lemma; I'll leave this post just in case someone wants to know the answer $$\ln(n+1) \leq \ln n + 1/n$$ $$\ln \frac{n+1}{n} \leq \frac{1}{n}$$ $$1+\frac{1}{n} \leq e^\frac{1}{n}$$ $$\frac{1}{n} \leq e^\frac{1}{n}-1$$ $$\frac{1}{n} \leq \sum_{k=0}^\infty\frac{1}{k!n^k}-1$$ $$\frac{1}{n} \leq 1+\frac{1}{n} + \sum_{k=2}^\infty\frac{1}{k!n^k}-1$$ $$0 \leq \sum_{k=2}^\infty\frac{1}{k!n^k}$$ Then, all you need to prove is that $\frac{1}{n^k k!}\geq 0$.