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How would I go about proving that $\lim_{n\to\infty}\sqrt{n+1}-\sqrt{n}=0$? I have tried to use Squeeze theorem but have not been able to come up with bounds that converge to zero. Additionally, I don't think that converting to polar is possible here.

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5 Answers 5

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$$ \sqrt{n+1}-\sqrt{n} = \frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} = \frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2\sqrt{n}} $$

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  • $\begingroup$ Shouldn't the last part be $\leq \frac{1}{2\sqrt{n+1}}$? $\endgroup$ Dec 20, 2015 at 1:57
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    $\begingroup$ @cheesyfluff That is not true since $2\sqrt{n+1} > \sqrt{n+1} + \sqrt{n}$.. $\endgroup$ Dec 20, 2015 at 2:00
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    $\begingroup$ Can you explain how you realized you're supposed to use conjugates? It's not really enlightening otherwise. $\endgroup$
    – user541686
    Dec 20, 2015 at 5:19
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    $\begingroup$ For squeezing, we could use from your results $$ \frac{1}{2\sqrt{n+1}}<\frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2\sqrt{n}}$$ $\endgroup$ Dec 20, 2015 at 5:30
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    $\begingroup$ @Mehrdad I try conjugates whenever I see differences of the form $\sqrt{a}-\sqrt{b}$ and that $a-b$ is simple. $\endgroup$ Dec 20, 2015 at 5:52
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Using conjugate multiplication can be quite useful in cases like that: $$\sqrt{n+1}-\sqrt{n}=(\sqrt{n+1}-\sqrt{n}){\sqrt{n+1}+\sqrt{n}\over \sqrt{n+1}+\sqrt{n}}={(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})\over \sqrt{n+1}+\sqrt{n}}={(n+1)-n\over \sqrt{n+1}+\sqrt{n}}={1\over \sqrt{n+1}+\sqrt{n}}$$. The last result is pretty easy to work with.

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One way is by using the mean value theorem. Specifically, let $f(x) = \sqrt x$. Then, for each $x > 0$, we know that $\displaystyle f(x+1) - f(x) = \frac{f(x+1) - f(x)}{(x+1) - x} = f'(c)$ for some $c$ in the interval $(x, x+1)$. Since $\displaystyle f'(x) = \frac1{2\sqrt x}$ is strictly decreasing we conclude that $0 < \displaystyle \sqrt{x+1} - \sqrt{x} < \frac1{2\sqrt x}$.

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Note that for positive $n$ we have $$\sqrt{n}\lt \sqrt{n+1}\lt \sqrt{n}+\frac{1}{2\sqrt{n}}.\tag{1}$$ The second inequality in (1) holds because $$\left(\sqrt{n}+\frac{1}{2\sqrt{n}}\right)^2=n+1+\frac{1}{4n}\gt n+1.$$ It follows from (1) that $$0\lt \sqrt{n+1}-\sqrt{n}\lt \frac{1}{2\sqrt{n}}.$$ Now Squeeze.

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I am gonna attempt to answer in a way that doesn't seem like pulling a rabbit outta the hat, and shows how I arrived at a solution.

It suffices to have that $(\sqrt{n+1} - \sqrt n)^2\to 0$, i.e., $2n + 1 - 2\sqrt{n^2 + n}\to 0$. Now, note that $2\sqrt{n^2 + n}$ is large enough to cancel $2n$, but then we also have to take care of the lingering $1$. This motivates us to take a nonnegative function $f$ such that $$ \sqrt{n^2 + n}\ge n + f(n)\tag{1} $$ so that $(\sqrt{n+1} - \sqrt n)^2\le 1- 2f(n)$. Thus it suffices to have that $$ \lim_{n\to\infty}f(n) = 1/2.\tag{2} $$

We thus find $f$ satisfying (1) and (2):

In view of (2), take $f$ to be the constant function $1/2$. But then, one quickly sees that this makes (1)'s RHS too large to be true. Thus we can try to reduce $f$, but in such a way that (2) still holds. Well, the first obvious thing that came to my mind was this: $$ f(n) := \frac{1}{2}\, \frac{n}{n + 1} $$ Now, it's straightforward to check that this indeed satisfies (1).

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