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Re: Sum of Infinite Series



John,

I'm afraid I didn't understand either of your replies. In your first one,
you suggest a Taylor series for - log(1 + x). What does this function have
to do with the series 1/2 + 1/4 + 1/6 + . . .?

In your second suggestion, you graph the hyperbola 1/(1 + x). Again, what
does this function have to do with the series 1/2 + 1/4 + 1/6 + . . .?

Paul O. Johnson

----- Original Message -----
From: "John S. Denker" <jsd@MONMOUTH.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Friday, January 25, 2002 5:11 AM
Subject: Re: Sum of Infinite Series


Paul O. Johnson wrote:

.... requires that I sum the infinite series 1/2 + 1/4 +
1/6 + 1/8 + . . .

Here's another way of looking at it that may be more
suited to the purpose. In particular, Paul's customers
may be less interested in the sum of the _infinite_
series than in the sum of a large but finite number
of terms.

So: Make a graph showing the hyperbola 1/(1+x). Then, on
the same graph, show the following stairstep function:
-- On the interval x= 0 to 1, y=1
-- On the interval x= 1 to 2, y=1/2
-- On the interval x= 2 to 3, y=1/3
-- On the interval x= 3 to 4, y=1/4

Now the area under the stairstep function is the series
of interest. The area under the hyperbola can be found
by integral calculus.

Note that the ordinates of the stairstep function are
"borrowed" from the value of the hyperbola at the left
edge of the interval, so the stairstep is strictly
above the hyperbola everywhere, so the sum of the
series is bounded below by the integral. Specifically,
the sum of the first N terms is bounded below by log(N+1).

Homework: By similar means you can construct an upper
bound on the series.

We now have upper and lower bounds on the partial sums,
so we have a pretty good idea how fast the series
diverges.

=================

You can get a much tighter bound by integrating the
function 1/(x+.5), which goes through the stairstep
function at the _middle_ of each unit interval.

Homework:
-- Is this integral an upper bound or a lower bound
on the series?
-- Prove it!
-- Construct a similar bound on the other side.


I get that the sum of the first N terms
S(N) = 1 + 1/2 + 1/3 ... 1/N
is approximated by (and bounded by)
log(N+.5) + log(2)
which is good to 10% when N=1 and gets better
and better (percentage-wise) for large N.

You can do even better by summing the first M
terms by hand, and using the integral trick to
estimate the remaining N-M terms.

Yet-better approximations are possible, but let's
stop here. This is the best I can do right now
using only relatively-elementary techniques with
nice relatively-accessible graphical representations.

Notice the relationship to the notion of Riemann
integral of the hyperbola: Multply the measure
of some interval by a "typical" ordinate on that
interval. The integral results from taking the
limit of really small intervals; the series results
from _not_ taking the limit, just using unit-sized
intervals.

===================

This illustrates a philosophical / pedagogical point
I've made before: In real-world physics, there are
lots of things that can't be solved exactly (or aren't
worth the trouble of solving exactly). So, very
often the job revolves around making a _controlled
approximation_.

Notice I said _controlled_ approximation. It's one
thing to have a wild-ass guess; it's quite another
to have rigorous upper and lower bounds.