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I'm working on finding whether sequences converge or diverge. If it converges, I need to find where it converges to.

From my understanding, to find whether a sequence converges, I simply have to find the limit of the function.

I'm having trouble getting started on this one (as well as one more, but I'll stick to one at a time).

I would appreciate if someone could explain how I should start this one.

ConfusingCalc
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  • Both seem to grow, but the square root grows quicker. Is it safe to assume that it goes to 0, or should I be using L'Hospital's rule to be checking this as another user mentioned? – ConfusingCalc Nov 28 '13 at 03:30
  • It is helpful to keep in mind that, at $+\infty$, $e^x$ is stronger than any polynomial or any $x^\alpha$ (positive exponent), which are in turn stronger than logs. This is often proved (essentially) with L'Hospital. – Julien Nov 28 '13 at 03:33
  • @julien Very interesting way to think about it. I wouldn't have ever pulled it apart and analyzed it like that. Thanks for the insight. – ConfusingCalc Nov 28 '13 at 03:38

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You need to use L'Hopital's rule, twice. What this rule states is that basically you if you are trying to take the limit of something that looks like $\frac{f(x)}{g(x)}$, you can keep on taking the derivative of the numerator and denominator until you get a simple form where the limit is obvious. Note: L'Hopitals rule doesn't always work.

We have: $$\lim_{n\to \infty} \frac{ln(n+1)}{\sqrt{n}} $$

We can keep on taking the derivative of the numerator and denominator to simplify this into a form where the limit is obvious:

$$= \lim_{n\to \infty} \frac{\frac{1}{n+1}}{\frac{1}{2\sqrt{n}}} $$ $$= \lim_{n\to \infty} \frac{(n+1)^{-1}}{(2\sqrt{n})^{-1}} $$ $$= \lim_{n\to \infty} \frac{2\sqrt{n}}{n+1} $$

Hrm - still not clear. Let's apply L'Hopitals rule one more time: $$= \lim_{n\to \infty} \frac{\frac{1}{\sqrt{n}}}{1} $$ $$= \lim_{n\to \infty} \frac{1}{\sqrt{n}} $$

The limit should now be obvious. It is now of the form $\frac{1}{\infty}$, which equals zero.

$$\lim_{n\to \infty} \frac{ln(n+1)}{\sqrt{n}} = 0 $$

Bob Shannon
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We use that $$\log x=\int_1^x t^{-1}dt$$

Let $\alpha>0$; choose $0<\varepsilon <\alpha$. Then $$\frac{\log x}{x^{\alpha}}=\frac{1}{x^{\alpha}}\int_1^x t^{-1}dt<\frac{1}{x^{\alpha}}\int_1^x t^{\varepsilon-1}dt<\frac{x^{\varepsilon-\alpha}}{\varepsilon}\to 0$$

Pedro
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    I like this proof because it is a direct use of definition of $\log x$ and it proves a very general result in the simplest possible way.(+1) – Paramanand Singh Nov 28 '13 at 05:00
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I assume you want the limit as $n \to \infty$. $\ln n$ grows more slowly than any positive power of $n$ as $n \to \infty$, including $\sqrt{n}$. The $+1$ in $\ln(n+1)$ is basically a distraction. So the answer is zero.

This is a little like killing a mosquito with a sledgehammer, but the conditions of L'Hopital's Rule are met (check them), you can apply it, and it works out nicely.

Stefan Smith
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  • Ah, so because ln(n+1) grows more slowly (even though it seems to constantly grow), it will go to 0 because $\sqrt{\infinity}$ explodes quicker? – ConfusingCalc Nov 28 '13 at 03:28
  • L'Hopital's rule is way too overestimated. Let's get over it already. – Pedro Nov 28 '13 at 03:30
  • @ConfusingCalc : $\ln(n+1)$ goes to $\infty$ extremely slowly as $n \to \infty$, much more slowly that $\sqrt{n}$ or any other positive power of $n$ (don't write "$\sqrt{\infty}$": at best it makes this confusing). L'Hopital's Rule is the quickest way I could think of proving the limit is $0$. – Stefan Smith Nov 28 '13 at 03:47
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    @PedroTamaroff Especially since L'Hospital did not even invent that rule, if I recall correctly. – Julien Nov 28 '13 at 03:48
  • @StefanSmith My comment aimed at the opposite direction. L'Hôpital is not such a sledgehammer, it is an effective tool, but simple. – Pedro Nov 28 '13 at 03:49
  • @StefanSmith I'm practicing for a quiz, so I'm assuming my professor will expect L'Hospital :) which is a good thing as my eyes are not near as good as you guys who immediately realized what the limit was! Thanks for the info – ConfusingCalc Nov 28 '13 at 03:56
  • @PedroTamaroff : sorry, I thought you were dissing L'Hopital's Rule. I have encountered people here who seem to have a phobia of it. It is indeed effective and simple. I used the "sledgehammer" cliche because it seems there ought to be a more basic way of evaluating this limit. – Stefan Smith Nov 28 '13 at 04:04
  • @julien : you probably know, but someone posted a long list of math theorems, lemmas, etc. on MSE that are named after the wrong people. It is huge. Appropriately, the list is named after someone who did not create it. I'm sorry I don't recall how to find it. – Stefan Smith Nov 28 '13 at 04:06
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    @StefanSmith Wikipedia does that too... – Julien Nov 28 '13 at 04:07
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    @StefanSmith Found it :-) – Julien Nov 28 '13 at 04:21
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Use L'Hospital's rule. Namely, if $\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}=L$, then $\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=L$. In your case just take in terms of x rather than n, so $f(x)=\ln(x+1)$ and $g(x)=\sqrt(x)$, then take the derivatives and find the limit.

Tristen
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  • So it is a L'Hospital case of infinity/infinity? – ConfusingCalc Nov 28 '13 at 03:33
  • Specifically, if we do assume that $n$ goes to infinity, since $f'(x)=\frac{1}{x+1}$ and $g'(x)=\frac{1}{2\sqrt(x)}$, you get $\lim_{x \rightarrow \infty}\frac{2\sqrt(x)}{x+1}=0$ – Tristen Nov 28 '13 at 03:41
  • Thank you for explaining the process! I knew I was missing something simply. Time to freshen up on L'Hospital and limits! Also, looks like we would have to apply L'Hospital again! – ConfusingCalc Nov 28 '13 at 03:44
  • No worries. I knew looking at it that it was zero, but L'Hospotials rule can often be the fastest way to show it! :) – Tristen Nov 28 '13 at 03:46
  • You could use it again if you want but it's really not necessary. Remember you just need to think in terms of powers to see its behavior most of the time. Just like $\frac{x}{\mathrm{e}^{x}}$ will converge pretty fast, so will $\frac{\sqrt(x)}{x}$. In your problem, just think about a number that's easy to calculate and you can see what it's going to do, like $x=100$. If you get in the habit of that it will make the process of finding limits much easier in general. – Tristen Nov 28 '13 at 12:28
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Let's consider the subsequence of the sequence $a_n = \frac{\ln(n+1)}{\sqrt{n}}$ where $n = e^m-1$. This subsequence is given by $$b_m = \frac{m}{\sqrt{e^m-1}}$$

We have that $$0 < b_m < \frac{m}{\sqrt{e^{m-1}}} \qquad\mbox{for } m \ge 2$$

and as $m\to\infty$, the right-hand side of this inequality goes to zero, so $(b_m)$ converges to $0$ by the Squeeze Theorem.

Of course, having a subsequence which converges does not mean a sequence converges in general, but it does for sequences which are eventually monotone increasing/decreasing. We'll show that $a_n$ is eventually monotone decreasing.

Let

$$f(x) = \frac{\ln(x+1)}{\sqrt{x}}$$

Then, $$f'(x) = \frac{\sqrt{x}\frac{1}{x+1}-\frac{1}{2\sqrt{x}}\ln(x+1)}{x} = \frac{2x - (x+1)\ln(x+1)}{2x^{\frac{3}{2}}(x+1)}$$

and for $x$ sufficiently large, say, $x>X$, $(x+1)\ln(x+1) > 2x$ (since $\ln(x+1)$ certainly grows larger than $2$ for $x$ large!), so the derivative will be negative for $x>X$ and $a_n$ will decrease for $n>X$. Since we've shown that a subsequence of $a_n$ converges, $a_n$ also converges to the same limit.