Expert: Scotto - 4/20/2009

Question
Hey Scotto,

I have a word problem pertaining to infinite series that I have no clue how to do. Here is the problem:

"The series ((-1)^(n+1))/n and 1/(n*2^n) both converge. Both their sums are equal to ln2. Why? Which series converges "faster" (and so numerically gives a more efficient way to get a numerical approximation for ln2)? Justify your answer by computing how many terms of each series must be added up to approximate ln2 with maximum allowed error of 10^-6."

It would be terrific if you could answer this problem and show all the steps. It would be great if I got the answer back by tomorrow. Thanks a lot Scotto. I really appreciate you taking the time to help.

Rob

Answer
I hope someone else could give you a solution to this.
I also pray you're having a good day.
There are two things I can think of to say,
and I also pray you'd get the solution somewhere.

It was thought it was a little strange when is says, 'Hi Scotto',
to me, but I also notice that the message was not private.
I mean, that's OK, but maybe someone else may answer it for you.
That's alright if the question is public,
since I can't give you an answer.

What I can say about the problem is this:
Given the two series above, ((-1)^(n+1))/n and 1/(n*2^n),
the second one must converge faster since the terms decrease faster.
I mean, since it's decreasing faster,
it must be converging on the point quicker.

One last thing to note:
Another way to look at this problem
is to say that we are trying to prove that 2 = π e^(Σ n_i),
where π is for the product of i and n_i is the term.

Another way I've seen them proved is by multiplying by some factor,
subtracting the original, and then cancelling terms in the sum.
I'm not sure this would work either, but just keep thinking about it.