Prove: $s_n \to s \implies \sqrt{s_n} \to \sqrt{s}$

Prove: $s_n \to s \implies \sqrt{s_n} \to \sqrt{s}$ by the definition of the limit. $s \geq 0$ and $s_n$ is a sequence of non-negative real numbers.

This is my preliminary computation:

$|\sqrt{s_n} - \sqrt{s}| < \epsilon$

multiply by the conjugate:

$|\dfrac{s_n - s}{\sqrt{s_n}+\sqrt{s}}| < \epsilon$

Thus we can use the fact that $|\sqrt{s_n} - \sqrt{s}| < \dfrac{|s_n - s|}{\sqrt{s}} < \epsilon$

After this I am lost...

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The condition $s\geq 0$ is not sufficient. You can have something like $S_n=-1/n$ and then the square root is never defined – Belgi Oct 3 '12 at 2:20
I just edited the question to clarify that $s_n$ is a sequence of non-negative real numbers. Thanks. – CodeKingPlusPlus Oct 3 '12 at 2:23
Hint: Do separate proofs for $s=0$ and $s\ne 0$. (In principle should also prove $s$ cannot be negative.) For $s=0$, from the $\epsilon$ it should not be hard to find appropriate $N$. For $s\gt 0$, your inequality at the end brings you close to finding an appropriate $N$. – André Nicolas Oct 3 '12 at 2:31
When doing $\epsilon-\delta$ proofs, you shouldn't start with $|\sqrt{s_n} - \sqrt{s}| < \epsilon$ statement. You should start with $|\sqrt{s_n} - \sqrt{s}| <$ something < ... < ... and work your way towards $\epsilon$. Otherwise, is very confusing and hard to work either on the left or right of the inequality. Hope I made myself clear. – Pragabhava Oct 3 '12 at 2:41
@Pragabhava I don't understand your proof. Could you write it out formally? – CodeKingPlusPlus Oct 3 '12 at 2:42

If both $s$ and $s_n$ are non-negative

$$|\sqrt{s}-\sqrt{s_n}|^2 \le |\sqrt{s}-\sqrt{s_n}||\sqrt{s} + \sqrt{s_n}|.$$

Step by Step :)

Since both $s$ and $s_n$ are non-negative

$$|\sqrt{s}-\sqrt{s_n}| \le |\sqrt{s} + \sqrt{s_n}|$$

this is clear because the result of substracting a non-negative number from another is always less than the result of adding it, then

$$|\sqrt{s}-\sqrt{s_n}|^2 \le |\sqrt{s}-\sqrt{s_n}| \cdot |\sqrt{s}+\sqrt{s_n}| = |s - s_n|$$

and you are done!

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I like that!! I see that $|\sqrt{s} - \sqrt{s_n}| \leq |\sqrt{s} - \sqrt{s_n}|^2 \leq |s - s_n|$. Now, this implies convergence because there is an $\epsilon$ s.t. $\forall_{n>N} |s-s_n| < \epsilon$ And therefore we can write: $|\sqrt{s} - \sqrt{s_n}| \leq |s - s_n| < \epsilon$ – CodeKingPlusPlus Oct 3 '12 at 3:05
@CodeKingPlusPlus Be carefull! Your first inequality does not hold if $|\sqrt{s} - \sqrt{s_n}| < 1$. What you say is $|\sqrt{s} - \sqrt{s_n}|^2 \le |s - s_n| \le \epsilon \Longrightarrow |\sqrt{s} - \sqrt{s_n}| \le \sqrt{\epsilon}$ – Pragabhava Oct 3 '12 at 3:10
How do we get $|\sqrt{s} - \sqrt{s_n}| \leq |s - s_n|$? Isn't that what we need? – CodeKingPlusPlus Oct 3 '12 at 3:27
@CodeKingPlusPlus No, what you need is to prove that $|s - s_n| < \epsilon$ implies that For all $\epsilon_1 > 0$ there exist an $N$ such that $|\sqrt{s} - \sqrt{s_n}| < \epsilon_1$ for $n > N$. Now, you've proven that such $N$ exist when $\epsilon_1 = \sqrt{\epsilon}$ – Pragabhava Oct 3 '12 at 3:47

$$\left| {\frac{{{s_n} - s}}{{\sqrt {{s_n}} + \sqrt s }}} \right| < \frac{{\left| {{s_n} - s} \right|}}{{\sqrt s }}$$

Since $s_n\to s$, for every $\epsilon >0$ there is an $n_0$ for wich $$\left| {{s_n} - s} \right| < \varepsilon \sqrt s$$ whenever $n\geq n_0$ (i.e. $\varepsilon \sqrt s$ is also an $\epsilon'>0$). Then, for this $n_0$, $$\left| {\frac{{{s_n} - s}}{{\sqrt {{s_n}} + \sqrt s }}} \right| < \frac{{\left| {{s_n} - s} \right|}}{{\sqrt s }} < \frac{{\varepsilon \sqrt s }}{{\sqrt s }} = \varepsilon$$

which means $\sqrt {s_n}\to\sqrt s$.

You're almost done. You arrived at

$$\left|\dfrac{s_n - s}{\sqrt{s_n}+\sqrt{s}}\right|$$

You know that $s_n\to s$, so you can make $|s_n-s|$ as small as you wish. Now, we need to know how to handle $\sqrt{s_n}+\sqrt{s}$. Since $s_n\to s$, there is an $n_0$ for wich

$$|s-s_n|<3s/4$$

Since $s_n>0$,$s\geq 0$, then

$$s-s_n\leq|s-s_n|<3s/4$$

This means that

$$s_n>s/4$$

then

$$2\sqrt {{s_n}} > \sqrt s$$

or, since $\sqrt s>0$

\eqalign{ & 2\sqrt {{s_n}} + 2\sqrt s > 3\sqrt s \cr & \sqrt {{s_n}} + \sqrt s > \frac{{3\sqrt s }}{2} \cr & \frac{1}{{\sqrt {{s_n}} + \sqrt s }} < \frac{2}{{3\sqrt s }} \cr}

Again, since $s_n\to s$, there is an $n_1$ for wich

$$|s-s_n|<\epsilon \frac{3\sqrt s}{{2 }}$$

Then, taking $n\geq \max\{n_0,n_1\}$ we have

$$\left| {\frac{{{s_n} - s}}{{\sqrt {{s_n}} + \sqrt s }}} \right| < \frac{2}{{3\sqrt s }}\left| {{s_n} - s} \right| < \frac{2}{{3\sqrt s }}\frac{{3\sqrt s }}{2}\varepsilon = \varepsilon$$

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Since $s_n\to s$, for any $\epsilon$, we can find an $N$ so that for all $n\ge N$ we have $|s_n-s|<\epsilon\sqrt{s}$. Then $$|\sqrt{s_n}-\sqrt{s}|=\left|\frac{s_n-s}{\sqrt{s_n}+\sqrt{s}}\right|\le\left|\frac{s_n-s}{\sqrt{s}}\right|$$ to get that $$|\sqrt{s_n}-\sqrt{s}|\le\epsilon$$

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well, you can actually take $\epsilon={\sqrt{s}}\epsilon'$ for any $\epsilon'>0$ given that for all $n>N, \epsilon'>|s_n-s|$

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It's the other way round: you're given $\epsilon$, then take $\epsilon'= \epsilon/\sqrt s$ and then an $N$ for $\epsilon'$. – lhf Oct 3 '12 at 2:51