2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – John Omielan Jan 14 at 2:13
  • 2
    $\begingroup$ Hint: consider $\exp(a_n)$ $\endgroup$ – Fakemistake Jan 14 at 2:17
  • $\begingroup$ This basically boils down to $\sqrt{x}$ growing faster than $\ln(x)$ for large $x$. $\endgroup$ – Rohit Pandey Jan 14 at 2:54
  • $\begingroup$ 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. $\endgroup$ – Aniruddh Venkatesan Jan 14 at 3:08
  • 1
    $\begingroup$ 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 :) $\endgroup$ – Aniruddh Venkatesan Jan 14 at 3:10
2
$\begingroup$

\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 $e^x$.

$\endgroup$
4
$\begingroup$

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 $x$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – stochasticboy321 Jan 14 at 3:55
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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