Limits of Sequences: 8.4 |s_n||t_n| < \frac{\epsilon}{M}

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Homework Help Overview

The discussion revolves around the convergence of sequences, specifically focusing on the inequality |s_n||t_n| < \frac{\epsilon}{M} as presented in a homework problem. Participants are examining the implications of this inequality in the context of limits of sequences.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the reasoning behind setting M equal to ε and the implications of that choice on the convergence definition. There is also a discussion on whether the inequality should hold strictly or if equality is acceptable.

Discussion Status

The discussion is active, with participants providing insights into the original paper's reasoning and questioning the author's notation and assumptions. Some participants are exploring different interpretations of the inequality and its requirements for convergence.

Contextual Notes

There is mention of a specific paper that may contain unclear or erroneous statements regarding the treatment of absolute values and inequalities in the context of convergence. Participants are reflecting on the clarity of the original problem statement and its implications for their understanding.

Artusartos
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In this link:

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

For number 8.4...

Why don't we just say...

[tex]|s_n||t_n| < \frac{\epsilon}{M} M = \epsilon[/tex]?

Thanks in advance
 
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If you mean, choose N0 such that if n> N0 then [itex]|s_n|< \frac{\epsilon}{M}[/itex] rather than M+1, that would, give us [itex]|s_nt_n|= \epsilon[/itex], not "<" which is required for the definition of convergence.
 
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HallsofIvy said:
If you mean, choose N0 such that if n> N0 then [itex]|s_n|< \frac{\epsilon}{M} rather than M+1, that would, as you say, give us [itex]|s_nt_n|= \epsilon[/itex], not "<" which is required for the definition of convergence.[/itex]
[itex] <br /> No I didn't say that...<br /> <br /> I said [tex]|s_nt_n| < \frac{\epsilon}{M} M = \epsilon[/tex]. So the first sign is an inequality.[/itex]
 
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.
 
Artusartos said:
Why don't we just say...

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.
 
Michael Redei said:
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.

Thanks :)
 

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