🚀 go-pugleaf

Brian,

Thanks for your post on Mike Bouldan's question about the integral of
1/x from -2 to 2. I found it quite helpful. Let me try to restate what I
*think* you're saying, and you tell me if I've got it.

-> The easiest example to understand is the infinite series
->
->       1 + -1 + 1 + -1 + 1 + -1 + ...
->
-> Using arguments not so very different from yours, we can either say
-> that this is the same as
->
->       (1 + -1) + (1 + -1) + (1 + -1) + ... = 0+0+0+... = 0
->
-> or say that the same series is equal to
->       1 + (-1 + 1) + (-1 + 1) + ... = 1+0+0+0+... = 1
->
-> Nothing is wrong with the manipulation in either case; the trouble is
-> that they don't prove "the answer is 0" or "the answer is 1."  They
-> prove "if there is an answer then it's 0" and "if there is an answer
-> then it's 1."  No contradiction there, since there isn't an answer;
-> the series diverges.

So, what you're saying is that any time a sum or difference involves a
divergent series, there is no answer to the sum or difference? Is this
true in all cases?

-> You can't say "infinity minus infinity is zero"; it could just as
-> well be any finite number (e.g., the number of counting numbers minus
-> the number of positive integers is 1), or infinite (e.g., the number
-> of integers minus the number of positive integers).

But there are techniques for evaluating limits that tend to infinity
minus infinity. These are commonly taught in first year calculus in the
chapter including L'Hopitals rule. How are these different from this
example?

I gotta make sure I know my stuff before I do this next year...

Sheila
coming to you from Diamond Bar, CA (in Los Angeles County)
                                                                                                                  

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