7
$\begingroup$

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.

$\endgroup$
20
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Shouldn't the last part be $\leq \frac{1}{2\sqrt{n+1}}$? $\endgroup$ – cheesyfluff Dec 20 '15 at 1:57
  • 3
    $\begingroup$ @cheesyfluff That is not true since $2\sqrt{n+1} > \sqrt{n+1} + \sqrt{n}$.. $\endgroup$ – Henricus V. Dec 20 '15 at 2:00
  • 2
    $\begingroup$ Can you explain how you realized you're supposed to use conjugates? It's not really enlightening otherwise. $\endgroup$ – Mehrdad Dec 20 '15 at 5:19
  • 1
    $\begingroup$ For squeezing, we could use from your results $$ \frac{1}{2\sqrt{n+1}}<\frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2\sqrt{n}}$$ $\endgroup$ – Claude Leibovici Dec 20 '15 at 5:30
  • 1
    $\begingroup$ @Mehrdad I try conjugates whenever I see differences of the form $\sqrt{a}-\sqrt{b}$ and that $a-b$ is simple. $\endgroup$ – Henricus V. Dec 20 '15 at 5:52
10
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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 $0 < \displaystyle \sqrt{x+1} - \sqrt{x} < \frac1{2\sqrt x}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.