From: Stephen Dalton 
Date: October 12, 2009 8:34:27 PM EDT
To: James Propp 
Subject: Comments on 92.305 Assignment #4

JIM: Assignment #4 is graded, but I won't get a chance to head up to UML to drop off the papers until Thursday as usual.  Here are the comments.

- A really important point: a sequence is not the same as a series.  An exercise that asks about a sequence generaly should not be solved using techniques intended to apply to series.  If the first line in a proof about a *sequence* is a statement about its "partial sums," then there's something amiss.  To be clear, a "sequence" is an ordered list of elements.  These elements are simply "listed," not added together or anything like that.  A "series," on the other hand, is a sum of elements.  A series is what you get when you add together the elements of a sequence.  Since it does not really make sense to tackle head-on the "addition of infinitely many things," a series is usually thought of as a limit of its partial sums.

This can be confusing; a series is not a sequence "in and of itself," but the partial sums DO form a sequence that has the same limit as the series.  To get a feel for what all of this means in context, look at the sequence - the SEQUENCE, not the series! - $a_n = 1/n^2$.  We could list the first few elements of this sequence:

a_1 = 1/1^2 = 1
a_2 = 1/2^2 = 1/4
a_3 = 1/3^2 = 1/9
...

And so on.  This "sequence" consists only and precisely of the infinite listing of these elements.  When we add the elements together, that's when we form a "series."  Note that just as the limit of a sequence cannot be found by locating the "last element" of the sequence (since there are infinitely many elements, there IS no "last element"), the infinite sum cannot be found by "adding all the terms together" (since adding infinitely many things together doesn't seem to make a lot of sense).  Instead, we form "partial sums," which are a sequence in and of themselves.  The first partial sum is just the first element of the sequence.  The second partial sum is the first element plus the second element; the third partial sum is the first element plus the second element plus the third element, and so on.  If we denote the partial sums by $s_n$, then we could write:

s_1 = a_1
s_2 = a_1 + a_2
s_3 = a_1 + a_2 + a_3
...
s_m = a_1 + a_2 + a_3 + ... + a_m

The infinite series, then, is defined to be the limit of these partial sums.  So in order to compute an infinite sum, we take the sum and turn it into a sequence, which we already know how to deal with!  But just to reiterate: calling something a "sequence" and calling something a "series" denote two very different things.  The relationship between them only begins when we turn the series into a "sequence of its partial sums."  When a question asks about a series, it is appropriate (and indeed, often necessary) to solve it by referring to the sequence of partial sums.  However, when a question asks about a SEQUENCE, most of the time it is not appropriate to turn it into a question about partial sums.

- Here's another important point.  A subsequence is an infinite listing of elements from the original sequence, in the same order, but with some elements "missing."  So if we have the sequence (1,2,3,...), a subsequence of it might be (1,3,5,...) or (2,4,6,...) or (2,3,5,7,11,...) (the prime numbers).  Technically, (1,2,3,...) is also considered a subsequence of itself, but that's not really all that useful.

A couple of things a subsequence is NOT:
(1) A subsequence is NOT a finite listing of the elements of the sequence.  The first ten, or twenty, or ten million elements of the sequence do not form a subsequence.
(2) A subsequence is NOT an algebraic "portion" of the sequence's general term.  If we have $b_n = (n^2+n)/n$, we can certainly simplify this to $b_n = n + 1$, but this does NOT indicate that $n$ and 1 are both subsequences of b_n.

- In Ex. 2.5.5, we are asked to produce the limit of the sequence $b^n$ if $-1 < b < 1$.  The case of $0 < b < 1$ was covered in Abbott, so we are supposed to focus on the case of $-1 < b \leq 0$.  Now, many students, after stating this part of the proof, more or less copied Abbott's proof that $b^n \rightarrow 0$ for $0 < b < 1$; they applied limit laws to the subsequence $b^{2n}$, and then solved algebraically for the limit.  The reason Abbott can do this in his proof is that he shows (using the Monotone Convergence Theorem) that the sequence $b^n$ converges for $0 < b < 1$ (it is bounded and monotone). He leaves the question of the value to which it converges open, but the fact that it converges allows Abbott to claim that the limit of the sequences is the same as the limit of all of its subsequences.  This step forms a vital basis for the rest of his proof, for it allows him to find the limit of the entire sequence by finding a limit of just one subsequence!  Many student solutions did the same sort of "limit law application" approach, but without first verifying that the sequence converged in the first place.  Reach a couple of weeks into your memory and you will recall that the limit laws apply only to *convergent* sequences, and so if you're going to apply the limit laws you really need to show first of all that the sequence in question converges.

- One student has said in an e-mail to me that he's trying hard to raise his grade after the past few homework assignments.  I understand that analysis can be tough, and sometimes especially tough for students who have done well in past courses.  This is a very different type of math than you may be accustomed to, and it can take a little while to get comfortable; this can pose particularly difficulty for students who seem to be hitting their first "intellectual ceiling" after getting by pretty easily in their courses up until now.  I just want to reassure everyone of two basic facts: (1) if you put in enough hard work, it WILL pay off and you WILL understand this stuff, and (2) the people who can put Ph.D after their name are not necessarily the people who understood everything the first time around, but mostly the people whose understanding is hard-won and the fruit of a lot of effort.  To quote Dr. Robert Borden in his Advanced Calculus, "Mathematics is beautiful, but it is crystal clear only to the genius."  I hope all of you are geniuses, but I know that I am not.  Being a genius is not the only way to do well in math, though.  Some students who may be discouraged by their performance thus far in the course could take heart in the fact that my own (eventual) success in analysis was not the result of any innate ability, but many, many long study sessions.  Keep trying, and keep rigorously applying the definitions you've learned in class and from the textbook, and it will pay off.

-Steve