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This says that if $\lim_{n\rightarrow\infty} a_n = l$ where $a_n$ and $l$ are reals, then $|\sqrt{a_n}-\sqrt\ell|\le \sqrt{|a_n-\ell|}$. I can't see how we can this inequality just by squaring and using triangular inequality.

As the other comment suggests, $(\sqrt{a_n} - \sqrt{l})^2 = (a_n - l) + 2 \sqrt{l}(\sqrt{l} - \sqrt{a_n}) < a_n - l$ has to use $a_n \ge l$. But how to derive this inequality without cases?

Silent
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  • Multiply the left hand side with $\sqrt{a_n} + \sqrt{\ell}$. – Daniel Fischer Aug 15 '17 at 15:47
  • You are saying in the question before that $\sqrt {a_n} + \sqrt l$ could be less than $1$. However, the fact that it is bounded below, since $a_n$ is a bounded sequence, shows that in fact $\sqrt {a_n} - \sqrt l$ is indeed always less than a constant multiple of $|a_n - l|$, which converges to zero. This avoids cases. – Sarvesh Ravichandran Iyer Aug 15 '17 at 15:48

3 Answers3

3

How about

$$\lvert \sqrt{a_n} - \sqrt{\ell}\rvert^2 \leqslant \lvert\sqrt{a_n} - \sqrt{\ell}\rvert\cdot(\sqrt{a_n} + \sqrt{\ell}) = \lvert (\sqrt{a_n} - \sqrt{\ell})(\sqrt{a_n} + \sqrt{\ell})\rvert = \lvert a_n - \ell\rvert,$$

which follows from $\lvert \sqrt{a_n} - \sqrt{\ell}\rvert \leqslant \sqrt{a_n} + \sqrt{\ell}$?

Daniel Fischer
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2

$$ \sqrt{|a|}=\sqrt{|a+b-b|} \leqslant \sqrt{|b-a|+|b|} \leqslant \sqrt{|b-a|}+ \sqrt{|b|}$$

$$ \sqrt{|b|}=\sqrt{|a+b-a|} \leqslant \sqrt{|b-a|+|a|} \leqslant \sqrt{|b-a|}+ \sqrt{|a|}$$

Thus $|\sqrt{|a|}- \sqrt{|b|}| \leqslant \sqrt{|a-b|}$

If $a_n,l \geqslant 0$ substitute and you can get rid of the absolute values.

Marios Gretsas
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Since $x\mapsto x^2$ is strictly increasing function on $[0,+\infty)$, we have

\begin{align} |\sqrt{a_n}-\sqrt l|\leq \sqrt{|a_n-l|} &\iff |\sqrt{a_n}-\sqrt l|^2\leq |a_n-l|\\ &\iff a_n+l-2\sqrt{a_n l}\leq |a_n-l| \end{align}

Assume $a_n\geq l$. Then

$$a_n+l-2\sqrt{a_nl}\leq a_n - l\iff l\leq \sqrt{a_nl} \iff l^2\leq a_nl\iff l\leq a_n.$$

By symmetry, we get $a_n+l-2\sqrt{a_nl}\leq l-a_n$ when $l>a_n$. Therefore, $$a_n+l-2\sqrt{a_nl}\leq |a_n-l|.$$

Ennar
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