# How did they find out what epsilon is equal to?

#### Plato

MHF Helper
For case I and case II, can you please explain how professor came to what epsilon was equal to?
https://imgur.com/a/BU09F6d
This may not help you at all. But I refuse to jump into someone else's proof.
We are given that $(\forall n)[S_n\ge 0]~\&~ S_n\to S$
1) We are to prove that $S\ge 0$ So let's suppose that $S<0$
Let $\varepsilon = \frac{{ - S}}{2}$. By convergence $(\exists N\in\mathbb{Z}^+)[n\ge N \Rightarrow |S - {S_n}| < \varepsilon$
Thus we have $\begin{gathered} \left| {{S_n} - S} \right| < \varepsilon \hfill \\ - \varepsilon < {S_n} - S < \varepsilon \hfill \\ {S_n} < \varepsilon + S < \frac{{ - S}}{2} + S = \frac{S}{2} < 0 \hfill \\ \end{gathered}$ But $S_N\ge 0$ so there is a contradiction.
Hence $S\ge 0$ as you were asked to prove for 1).
Now can you construct a proof for $\sqrt{S_n}\to\sqrt{S}~?$

2 people

#### math951

I'll get back to you in a couple hours on the 2nd part of the proof. Currently about to drive home.

#### math951

1 question from result..I understand it and makes sense, however, Why did you put absolute value of Sn-S is less than espilon and absolute value S-Sn is less than espilon as well? I don't see that in any of my convergent definitions. Why is that a legal thing we can do?

Moreover, to prove: $$\displaystyle \sqrt{S_n}\to\sqrt{S}$$

Proof: we are given $$\displaystyle (S_{n})\geq0(\forall n)\& S_{n}\to S$$
WTS: $$\displaystyle \sqrt{S_n}\to\sqrt{S}$$

$$\displaystyle (\exists N\in\mathbb{Z}^+)[n\ge N \Rightarrow |{S_n - S| < \varepsilon$$

$$\displaystyle \sqrt{S_n}-\sqrt{S}=\frac{\|S_n - S|{\sqrt{S_n}+\sqrt{S}} \Leftrightarrow \frac{\\epsilon}{\sqrt{S}+\sqrt{S}$$

So $$\displaystyle |\sqrt{S_n}-\sqrt{S}| <= \frac{\\epsilon}{2\sqrt{S}} (\forall n) >= N.$$ Hence proved by def. of limit.

Not sure why it is showing \displaystyle for 2 of my lines of TeX...I am using $$\displaystyle ....$$ Anyone know why?

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#### Plato

MHF Helper
1 question from result..I understand it and makes sense, however, Why did you put absolute value of Sn-S is less than espilon and absolute value S-Sn is less than espilon as well? I don't see that in any of my convergent definitions. Why is that a legal thing we can do?

Moreover, to prove: $$\displaystyle \sqrt{S_n}\to\sqrt{S}$$

Proof: we are given $$\displaystyle (S_{n})\geq0(\forall n)\& S_{n}\to S$$
WTS: $$\displaystyle \sqrt{S_n}\to\sqrt{S}$$

$(\exists N\in\mathbb{Z}^+)[n\ge N \Rightarrow |{S_n - S| < \varepsilon$