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Show that any null sequence $(a_n)_{n\in\mathbb R}, \ a_n\neq 0$ fulfils

$$\lim_{n\to\infty}\frac{\sqrt{1+a_n}-1}{a_n}=\frac{1}{2}\tag{1}$$

Was wondering if my approach is valid.

Let's assume that $\lim_{n\to\infty}a_n$ exists and that $\lim_{n\to\infty}=0$ then we know that $\lim_{n\to\infty} (\sqrt{1+a_n} -1)$ also exists. So we can write:

$$\lim_{n\to\infty}\frac{\sqrt{1+a_n}-1}{a_n}=\frac{\lim_{n\to\infty}\sqrt{1+a_n}-1}{\lim_{n\to\infty}a_n}=\frac{1}{2}\tag{2}$$

Now we know that we can rewrite this to

$$\lim_{n\to\infty}\sqrt{1+a_n}-1 = \frac{1}{2}\lim_{n\to\infty}a_n\tag{3}$$

We use the fact that the square root is continuous:

$$\lim_{n\to\infty}\sqrt{1+a_n}-1=\sqrt{\lim_{n\to\infty}1+a_n}-1 = \frac{1}{2}\lim_{n\to\infty}a_n\tag{4}$$

We get

$$\sqrt{1+0}-1=\frac{1}{2}\cdot 0\tag{5}$$

So the equation does hold for any null sequence $a_n$.

Question: Is there any flaw here?

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  • $\begingroup$ There must be a flaw, because using the same technique you could prove that $\lim_{n\to\infty}\frac{\sqrt{1+a_n}-1}{a_n} $ is $17$ or $\pi$ or whatever you want. $\endgroup$ – Martin R Jan 24 '19 at 14:47
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Regarding your question whether there is a flaw: You need to be careful with mixing the assumptions and what you actually want to prove. You cannot use $$\lim_{n\to\infty}\frac{\sqrt{1+a_n}-1}{a_n}=\frac{1}{2}$$ in (2) because this is what you actually want to prove! Even if the following steps are equivalences (and you have to be very careful there, in particular with square roots), it is confusing at best and therefore not a very solid proof. Therefore my advice: Try writing proofs in a way that you start with the assumptions and end up with the result instead of mixing the result in somewhere in the middle and conclude with a tautology. This will save you from a lot of trouble and also makes your proofs much easier to read.

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  • $\begingroup$ Yeah, I know I could prove it way more elegant. While it is clear that I can't use a claim to prove that claim, I'm not sure if that's really the case here. Could you elaborate? $\endgroup$ – xotix Jan 24 '19 at 12:45
  • $\begingroup$ Ok, to be more specific: In (2) you are using the claim you want to prove. Then you proceed to (3), (4) and so on and then end up with a true statement. For this to be consistent, all your steps need to be equivalences. This is very susceptible to mistakes as we are usually thinking top-down and you are reversing it here. In this particular case, there is a problem from (3) to (2) as you are dividing by $\lim_{n\to\infty}a_n$, which is assumed to be $0$. $\endgroup$ – Klaus Jan 24 '19 at 12:54
  • $\begingroup$ I think you meant multiplied, no? Anyway, I see your point. Thanks a lot. $\endgroup$ – xotix Jan 25 '19 at 13:39
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You divide by zero, which is not allowed. If you know a bit of calculus you can find the solution by examining the derivative of the root function at the point x=1.

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  • $\begingroup$ It's not zero, is it? It's a null sequence, which does kind of have an inverse, no? Not sure if I can call it inverse though. $\endgroup$ – xotix Jan 24 '19 at 12:41
  • $\begingroup$ You are dividing by the limit, which is zero. Diving a real number by a sequence wouldn't make any sense anyway. $\endgroup$ – Jagol95 Jan 24 '19 at 18:33
  • $\begingroup$ do you know basic single variable calculus? $\endgroup$ – Jagol95 Jan 24 '19 at 18:45
  • $\begingroup$ The solution is constructed to address question like that exactly. When I see stuff like this I start thinking what it exactly means and just treating a null sequence as the actual number 0 seems strange. Don't we actually work in an extended real space where $\pm \infty$ are symbol so we can work with it, e.g. we can now write $1/0=\infty$ etc.? But I see that this won't allow me what I did. $\endgroup$ – xotix Jan 25 '19 at 13:37
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Multiply numerator and denominator by $$\sqrt{1+a_n}+1$$

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  • $\begingroup$ That is most times called "the conjugate of the numerator (or denominator)" +1 $\endgroup$ – DonAntonio Jan 24 '19 at 12:05
  • $\begingroup$ Ok, that is shorter, the conjugate. $\endgroup$ – Dr. Sonnhard Graubner Jan 24 '19 at 12:06
  • $\begingroup$ i know these words only in connection with complex numbers $\endgroup$ – Dr. Sonnhard Graubner Jan 24 '19 at 12:13
  • $\begingroup$ The meaning is pretty similar: change the sign of the second term...that's all. $\endgroup$ – DonAntonio Jan 24 '19 at 12:42
  • $\begingroup$ How does this answer the “Question: Is there any flaw here?” $\endgroup$ – Martin R Jan 24 '19 at 14:30

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