# Prove $\frac{\ln{n}}{\sqrt{n+1}} > \frac{\ln{(n+1)}}{\sqrt{n+2}}$

Prove sequence $$a_n = \frac{\ln{n}}{\sqrt{n+1}}$$ is strictly decreasing for big $$n$$ (using elementary methods). Although I can use different methods to get this result, I would like to have this done without derivatives, etc, just basic definitions, limits, inequalities with $$\exp$$ and $$\ln$$.

EDIT: Details have been added to the question. Rohit Pandey's answer was addressing previous question statement.

• What do you specifically mean by "elementarily", in particular what it includes and/or excludes as methods permitted to be used. Also, please provide some context for your question, such as where it comes from, what you've tried, etc. Thanks. – John Omielan Jan 14 at 2:13
• Hint: consider $\exp(a_n)$ – Fakemistake Jan 14 at 2:17
• This basically boils down to $\sqrt{x}$ growing faster than $\ln(x)$ for large $x$. – Rohit Pandey Jan 14 at 2:54
• Here's a weird suggestion I'm not sure will work: You can show that the series $\sum_{n=1}^\infty{\frac{\ln(n)}{\sqrt{n+1}}}$ converges, and by the convergence tests, your identity must hold. – Aniruddh Venkatesan Jan 14 at 3:08
• Just read your edit, while the method of convergence tests are taught in Calc BC or Calc II, depending on your location, the actual method I am referring to requires no derivatives or integrals etc, just limits and a little bit of intuition :) – Aniruddh Venkatesan Jan 14 at 3:10

\begin{align} \frac{\log{n}}{\sqrt{n+1}} &> \frac{\log{(n+1)}}{\sqrt{n+2}} \\ \iff \sqrt{1 + \frac{1}{n+1}} \log n &> \log(n+1) \end{align}

Now $$\sqrt{1 + x} > 1 + x/3$$ for $$x < 1,$$ (this arises since I know $$\sqrt{1+x} \approx 1 + x/2$$ for small $$x$$, and then an elementary proof by factoring a quadratic is easy if I replace the $$2$$ with something bigger (like $$3$$) )

so it suffices to show that $$\log n + \frac{\log n}{3(n+1)} > \log (n+1) \iff \frac{\log n}{3(n+1)} > \log (1 + \frac{1}{n})$$

Further, since $$\log (1 + x) < x,$$ it suffices to show that

$$\frac{\log n}{3(n+1)} > \frac{1}{n}$$

But this is true for large $$n$$ - for $$n \ge 1,$$ $$3(n+1)/n \le 6$$, and picking $$n > e^6$$ does the job.

It remains to argue that $$\log(1 + x) < x$$ is elementary. The simplest way I know is to note that this is equivalent to $$1 + x < e^x,$$ which is true for $$x >0$$ by simply truncating the series definition of $$e^x$$.

Not sure what you mean by elementary methods, but one can simply take the derivative of $$f(x) = \frac{\ln(x)}{\sqrt{x+1}}$$. We get:

$$f'(x) = \frac{1}{\sqrt{x+1}}\left(\frac{1}{x} - \frac{\ln(x)}{2(x+1)}\right)$$

Since $$2(x+1) for large $$x$$, the derivative is negative and so $$f(x)>f(x+1)$$ should hold for large $$x$$.

It is (relatively) straightforward to show that

$${\ln n\over\sqrt{n+1}}\gt{\ln(n+1)\over\sqrt{n+2}}\iff{\ln n\over(\sqrt{n+1}+\sqrt{n+2})\sqrt{n+1}}\gt\ln\left(1+{1\over n} \right)$$

It is also easy to see, for example, that $$(\sqrt{n+1}+\sqrt{n+2})\sqrt{n+1}\lt(2\sqrt n+2\sqrt n)(2\sqrt n)=8n$$. Now if we take the inequality $$(1+{1\over n})^n\lt e$$ for granted, then, for $$n\gt e^8$$, we have

$$\ln\left(1+{1\over n} \right)={1\over n}\ln\left(\left(1+{1\over n} \right)^n\right)\lt{1\over n}\ln e={1\over n}\lt{\ln n\over 8n}\lt{\ln n\over(\sqrt{n+1}+\sqrt{n+2})\sqrt{n+1}}$$

Remark: the $$8$$ in the inequality $$(\sqrt{n+1}+\sqrt{n+2})\sqrt{n+1}\lt8n$$ is rather crude, but the OP only asked for a proof for "big" $$n$$. It might be interesting to see an elementary (non-calculus) answer that pegs more precisely where the OP's inequality kicks in.

• Interesting remark. I used a calculator for this, but - playing with the constants in my answer (which is morally the same as yours) lets it kick in at $n = 12$ by using $\sqrt{1+1/{n+1}} > 1 + (2.2076 (n+1))^{-1}$ which holds for $n \ge 12$ and then $\exp(2.2076 \times (13/12)) < 11.$ Probably this can be improved to 11. W/o a calculator I got values of around 20 - the main issue is upper bounding $\exp(2.a \times 1.b)$ accurately. That said the $\log x > 2(x+1)/x$ only starts holding at $9.2$ish anyway, so this is already super close. – stochasticboy321 Jan 14 at 3:55

In order to show that $$n\to \frac{\log n}{\sqrt{n+1}}$$ is decreasing from some point on, it is enough to show that $$x\mapsto\frac{x}{\sqrt{e^x+1}}$$ or $$x\mapsto\frac{x^2}{e^x+1}$$ are decreasing from some point on, or that $$x\mapsto \frac{e^x+1}{x^2}$$ is increasing from some point on. On the other hand all the derivatives of $$f(x)=\frac{e^x-1-x}{x^2}$$ are positive on $$\mathbb{R}^+$$ (just think to the Maclaurin series) and $$g(x)=\frac{x+2}{x^2}$$ is such that both $$g$$ and $$g'$$ converge to zero as $$x\to +\infty$$: it follows that $$f+g$$ is increasing from some point on.