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Consider the sequence $a_n = \sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\\$. To determine the limit I did the following: \begin{aligned} a_{n} &=\left(\sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\right) \frac{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}} \\[10pt] &=\frac{2 \sqrt{n}}{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}=\frac{2 \sqrt{n}}{\sqrt{n} \sqrt{1+\frac{1}{\sqrt{n}}}+\sqrt{n} \sqrt{1-\frac{1}{\sqrt{n}}}} \\[10pt] &=\frac{2}{\sqrt{1+\underbrace{\frac{1}{\sqrt{n}}}_{\rightarrow0 \text{ for }n \rightarrow \infty}}+\sqrt{1-\underbrace{\frac{1}{\sqrt{n}}}_{\rightarrow0 \text{ for }n \rightarrow \infty}}} \\[10pt] &= \dfrac{2}{2} = 1. \end{aligned}

However, my first thought was $a_n = \sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}} = \sqrt{n}\left(\sqrt{1 + \dfrac{1}{\sqrt{n}}}\space - \sqrt{1 - \dfrac{1}{\sqrt{n}}}\right) = \sqrt{n}\space (1-1) = 0$

with the same argument as above. Is it correct that one can't make a statement about the convergence in the latter calculation because we have $\infty \cdot 0$ ? If yes, why exactly is this the case?

Edit: To see that $\lim\limits_{n\to \infty} \dfrac{1}{\sqrt{n}} = 0$ pick some arbitrary $\epsilon > 0$. We want to find a $N$ s.t. $\forall n\geq N\colon |\dfrac{1}{\sqrt{n}} - 0|< \epsilon \Longleftrightarrow \dfrac{1}{\sqrt{n}} < \epsilon \Longleftrightarrow n >\dfrac{1}{\epsilon^2}$ meaning we can choose $N = \dfrac{1}{\epsilon^2} + 1$ for example.

saulspatz
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Hilberto1
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  • $0\cdot\infty$ could be any thing, since it all depends on how fast we are approaching $\infty$ vs how fast we are approaching $0$. So when you reach this case, you try a new method. – Rushabh Mehta Feb 28 '21 at 22:24
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    There seems to be an extra $\sqrt{}$ sign in the denominator of some of the fractions, but it's ignored sometimes. Going from the next-to-last line to the last line of the calculation, we get $\sqrt2$ in the denominator, but then, the $sqrt$ should be there. – saulspatz Feb 28 '21 at 22:24
  • Your first thought only works if you can show $\sqrt{1 + \frac{1}{\sqrt{n}}}\space - \sqrt{1 - \frac{1}{\sqrt{n}}}$ is $o\left(\frac1{\sqrt{n}}\right)$ which it is but you have not shown it. – Henry Feb 28 '21 at 22:26
  • Looks like the limit is $1$, but you've got some typos on the way, as pointed out in the other comments. – mjw Feb 28 '21 at 22:26
  • I think I've corrected the typos in the denominators, Please check that I haven't inadvertently changed what you meant to say. – saulspatz Feb 28 '21 at 22:41
  • The first calculation looks correct to me now. – saulspatz Feb 28 '21 at 22:42

2 Answers2

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When one factor of a product goes to $\infty$ and the other goes to 0, you can't immediately conclude what the limit is. The easiest way to see this is to consider $\lim_{n \to \infty} n \cdot \frac{1}{n}$. Now replace the 1 by any other number.

The error in your calculation is that when you evaluate a limit by substitution, you have to substitute all instances of $n$ at the same time. In your calculation, in the second to last = sign, you substitute some instances to get $1-1$ while leaving the $\sqrt{n}$ unsubstituted. That is wrong because when you take a limit, all instances of $n$ are tending to the limit at the same time.

Ted
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Your second approach is wrong.
To save writing let $x=\sqrt{n}$, and expand square roots in power series in $\frac{1}{x}$, so expression becomes $x(\sqrt{1+\frac{1}{x}}-\sqrt{1-\frac{1}{x}})$

$=x(1+\frac{1}{2x}+O(\frac{1}{x^2})-1+\frac{1}{2x}+O(\frac{1}{x^2}))$

$=1+O(\frac{1}{x})\to 1$ as $x\to \infty$

herb steinberg
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