# Limits of sequences

1. Dec 14, 2012

### Artusartos

http://people.ischool.berkeley.edu/~johnsonb/Welcome_files/104/104hw3sum06.pdf [Broken]

For number 8.4...

Why don't we just say...

$$|s_n||t_n| < \frac{\epsilon}{M} M = \epsilon$$?

Last edited by a moderator: May 6, 2017
2. Dec 14, 2012

### HallsofIvy

Staff Emeritus
If you mean, choose N0 such that if n> N0 then $|s_n|< \frac{\epsilon}{M}$ rather than M+1, that would, give us $|s_nt_n|= \epsilon$, not "<" which is required for the definition of convergence.

Last edited: Dec 14, 2012
3. Dec 14, 2012

### Artusartos

No I didn't say that...

I said $$|s_nt_n| < \frac{\epsilon}{M} M = \epsilon$$. So the first sign is an inequality.

4. Dec 14, 2012

### HallsofIvy

Staff Emeritus
Yes, you are right about what you said and I have edited my post to remove "as you said". But you are incorrect that it would be "<". You would have, instead, "=", as I said.

5. Dec 14, 2012

### Michael Redei

Probably because "we" didn't think that closely when "we" wrote that paper. If you ensure that $|s_n|<\frac\epsilon M$ what you actually get is
$$|s_nt_n| = |s_n|\cdot|t_n| < \frac\epsilon M\cdot|t_n| \leq \frac\epsilon M\cdot M = \epsilon,$$
which is what you have yourself.

If you look at that paper again, you'll see that the author writes
$$\begin{eqnarray*} |s_nt_n − 0| & = & |s_n| \cdot |t_n| \\ & < & \left|\frac\epsilon{M + 1}\right| \cdot |M| \\ & < & \epsilon. \end{eqnarray*}$$
Why did he suddenly need absolute values in the middle line? I think he probably didn't proofread what he'd written.

6. Dec 14, 2012

Thanks :)