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I am given this sequence with square root. $a_n:=\sqrt{n+1000}-\sqrt{n}$. I have read that sequence converges to $0$, if $n \rightarrow \infty$. Then I said, well, it may be because $\sqrt{n}$ goes to $\infty$, and then $\infty - \infty = 0$. Am I right? If I am right, why am I right? I mean, how can something like $\infty - \infty = 0$ happen, since the first $\infty$ which comes from $\sqrt{n+1000}$ is definitely bigger than $\sqrt{n}.$?

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    $\begingroup$ Would you say that the limit of $b_n:=(n+1)-n$ is 0 for the same reason? $\endgroup$
    – Charles
    Commented Jan 2, 2013 at 19:36
  • $\begingroup$ now i got, Charles. thanks a bunch $\endgroup$
    – doniyor
    Commented Jan 2, 2013 at 19:43
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    $\begingroup$ @doniyor The key thing is to remember that operations that we learn only hold for the integers, then the rationals, then the reals (and finally the complex numbers). Each time, when we extend it, we have to check that the definition makes sense (e.g. that we indeed have equivalence classes in fractions). In the case of trying to extend arithmetic to $\infty$, we have $\infty + * = \infty$, where $*$ denotes anything that we're adding to. Hence, $\infty - \infty$ could be any value, hence is undefined without knowing more. Same for $\frac {\infty}{\infty} = *$. $\endgroup$
    – Calvin Lin
    Commented Jan 2, 2013 at 21:41
  • $\begingroup$ yeah, you are totally right Calvin. thanks $\endgroup$
    – doniyor
    Commented Jan 3, 2013 at 16:54

7 Answers 7

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The argument $\infty-\infty=0$ will fail you if you consider $a_n=(n+1)-n$ where the limit is obviously $1$ or $n^2-n$ where the limit is $\infty$. Indeed if $a_n\to \infty$ and $b_n\to \infty$ you can say nothing for the covergence of $a_n-b_n$ (unless you have additional clues).

In our case, although $\sqrt{n+a}$ is strictly bigger than $\sqrt{n}$, both diverge to $+\infty$ in exactly the same way. In fact, $$\sqrt{n+a}-\sqrt{n}=\frac{n+a-n}{\sqrt{n+a}+\sqrt{n}}\to a\frac{1}{\infty}=0$$ for any $a\in \mathbb{R}$. For the first equality I used the well known identity: $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$$

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  • $\begingroup$ wow, okay, thanks in tons. i had bad feeling for infinity all the time $\endgroup$
    – doniyor
    Commented Jan 2, 2013 at 19:46
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    $\begingroup$ I know I will need this answer for my own class.+1000 $\endgroup$
    – Mikasa
    Commented Jan 3, 2013 at 18:29
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No, $\infty-\infty$ is indeterminate.

This is what you can do:

$$ \sqrt{n+1000}-\sqrt{n}=\frac{(\sqrt{n+1000}-\sqrt{n})(\sqrt{n+1000}+\sqrt{n})}{\sqrt{n+1000}+\sqrt{n}}=\frac{n+1000-n}{\sqrt{n+1000}+\sqrt{n}}. $$

Now you surely can see why this converges to $0$.

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  • $\begingroup$ thanks julien, now i got the clue $\endgroup$
    – doniyor
    Commented Jan 2, 2013 at 19:40
  • $\begingroup$ @doniyor, julien wasn't just giving you a clue. What you have is an indeterminate form, and the general scheme of things is to turn to l'hopital's rule for such problems. $\endgroup$ Commented Jan 2, 2013 at 19:43
  • $\begingroup$ julien showed how to come to that $0$-limit. and tought me that my thought is completely wrong $\endgroup$
    – doniyor
    Commented Jan 2, 2013 at 19:44
  • $\begingroup$ Ok just wanted you to know if something similar came up. $\endgroup$ Commented Jan 2, 2013 at 19:47
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But this is not really $\infty-\infty$. This is the limit of differences between two unbounded sequences. This is the key point here.

The difference is approaching zero, because $1{,}000$ is very small compared to $\sqrt n$. At some point, $\sqrt n$ is almost the size of $\sqrt{n+1000}$. For example for $n=10^{1000}$ the difference gets very small, that is $a_{10^{1000}}$ is a very small number.

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You have to look at the sequence in particular. In your case, a little algebra clears things up:

$$ \sqrt{n+1000} - \sqrt{n} = \frac{1000}{\sqrt{n+1000} + \sqrt{n}} \approx \frac{1000}{2 \sqrt{n}} $$

which obviously goes to zero as $n \rightarrow \infty$.

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$$\sqrt{n+100}-\sqrt n=\frac{100}{\sqrt{n+100}+\sqrt n}\xrightarrow [n\to\infty]{}0$$

But you're not right, since for example

$$\sqrt n-\sqrt\frac{n}{2}=\frac{\frac{n}{2}}{\sqrt n+\sqrt\frac{n}{2}}\xrightarrow [n\to\infty]{}\infty$$

In fact, "a difference $\,\infty-\infty\,$ in limits theory can be anything

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Note that $$\sqrt{n}\lt \sqrt{n+1000}\lt \sqrt{n}+\frac{500}{\sqrt{n}}.$$ This is because $$\left(\sqrt{n}+\frac{500}{\sqrt{n}}\right)^2=n+1000+\frac{500^2}{n}\gt n+1000.$$

If follows that $$0\lt \sqrt{n+1000}-\sqrt{n}\lt \frac{500}{\sqrt{n}}.$$ Now by Squeezing we can conclude that $$\lim_{n\to\infty}(\sqrt{n+1000}-\sqrt{n})=0.$$

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I think you need to revise your definition of limit of a sequence.

$\sqrt{n+1000}$ comes arbitrarily close to $\sqrt{n}$. More precisely, for any arbitrarily small $\epsilon > 0$ there exists an integer $N_\epsilon$ such that,

\begin{equation} \sqrt{n+1000}-\sqrt{n} < \epsilon \;\; \forall \; n>N_\epsilon \end{equation}

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