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

Question

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.

Answer 1

\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 $e^x$.

Answer 2

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

Answer 3

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.

Answer 4

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.