From: Stephen Dalton
Date: October 8, 2009 10:39:16 PM EDT
To: James Propp 
Subject: Comments on assignment #3

Hello 92.305 Students,

Some comments from me, your humble grader, on assignment #3.

- Some students gave what I would rule incomplete answers on 2.2.1.  They argued in the same style as Abbott in Example 2.2.6, where his last line states that $1/n < \epsilon$ and hence we can conclude that $\left| \frac{n+1}{n} - 1 \right| < \epsilon$.  Now, this is not *too* difficult to see algebraically, but look a few lines above to see that Abbott has *shown* in a previous step that we may obtain the second inequality from the first.  It seems to me a little hurried to claim two inequalities are equivalent without going through any grunt-work to show it.  For example, In 2.2.1.a, something like the equivalence of the inequality $n > \sqrt{ \frac{1/\epsilon - 1}{6}}$ and the inequality $\frac{1}{6n^2} < \epsilon$ is a bold enough claim, at least in my mind, to need evidence.  It really is necessary to mold the $n > N$ inequality all the way into precisely the $|a_n - a| < \epsilon$ inequality.  If you do it by means of Abbott's method in 2.2.6 (i.e., accomplishing the last few steps by referring to earlier work) then that's absolutely fine, but please justify the leap in some way.  ***I did not penalize students who made this error for this assignment, since it is the first assignment that involves computing and proving limits.***

- Problem 2.3.1.  A sequence being "constant" does not permit us to employ Theorem 2.3.3, which speaks of what happens when we multiply a sequence by a constant.  A sequence being "constant," as is stated in the problem, means that all of its elements have the same value.  Some examples of constant sequences would be (1, 1, 1, ...) or (9.6, 9.6, 9.6, ...).  This is not the same concept as multiplying a sequence by a constant!

- Problem 2.3.5.  There were some issues with this one.  Again, Theorems 2.3.3 and 2.3.4 do not apply to this.  Also, lim{x_n} = lim{y_n} does not imply that x_n and y_n are "the same."  For example, if we take x_n = 1/n and y_n = -1/n, the limits are the same, but the sequences do not share any members at all!  I highly suggest everyone check out the web solution.

Also, a couple of brave students might try a crack at this alternate way of showing that z_n converges based on the convergence of x_n and y_n.  The class is covering subsequences right now, right?  Well, let's say that we have a sequence (z_n) - maybe convergent, maybe divergent, we don't know yet - and we split it into two subsequences (z_{n_i}) and (z_{n_j}).  If the n_i's and the n_j's cover all of the natural numbers - that is to say, if these two subsequences, between them, contain all the members of (z_n) - and if *both of the subsequences converge to the same limit*, then try to show that (z_n) converges, and to the same limit as the subsequences.  (This can be applied nicely to the (z_n) defined in 2.3.5.  It could even be applied if we made the "shuffling" weirder, like (x_1,x_2,y_1,x_3,x_4,y_2,...)!  Sometimes problems get simpler if you prove preliminary results first.)

- General note: I would advise students to assume that if they need to show something is true, they need to produce a proof, and if they need to show something is NOT true, they need to show a counterexample.  Sometimes Abbott does say explicitly that proofs are not required and that we should try to "guess" whether a statement seems true or false based on our intuition, but if this is not explicitly stated then it is only appropriate to claim something true if one has proven it to be true and false if one has produced a counterexample.

- Another general note.  Just because a sequence is bounded does not mean that it converges!  This is very important.  A convergent sequence must be bounded, ***but the converse is not necessarily true***.  There are loads of bounded sequences that do not converge.  Take (1, -1, 1, -1, ...), for example.  Theorem 2.3.2 in the text is NOT meant to indicate that every bounded sequence converges, but ONLY that every convergent sequence is bounded!

- A LaTeX hint!  If you want to put something in quotes in LaTeX, you cannot just put them in normal quotes like you do with MS Word.  LaTeX doesn't read the quotation marks correctly; with Word, it interprets the first quotation mark you put down as an opening quotation mark, and makes the second one a closing one.  LaTeX doesn't do this automatically.  In order to get a left-hand quotation mark (the one you put at the beginning of a quote), use two consecutive copies of the ` key (directly left of the key for the number 1 on most keyboards).  The closing quote can just be the normal quotation mark next to the Enter key that you're used to using from Word.  ``So LaTeX would typeset this correctly,'', whereas "it would not typeset this correctly."

-Steve