Associated Topics || Dr. Math Home || Search Dr. Math

### Sum of Any Infinite Series

```
Date: 11/19/2000 at 02:00:28
From: Dan Keesing
Subject: If an Infinite Series Converges

Hi Dr. Math,

I just learned about various tests for convergence and the "powerful"
Taylor series, but I was disappointed to learn that even if you can
determine that a series converges, it's usually impossible to
calculate an exact answer for the sum. Except for the rare cases like

inf.
SUM (1/n^2) = pi^2/6
n=1

by Euler's proof, it would be extremely difficult to find an exact
answer for an infinite sum. I am having trouble understanding the idea
that if an infinite sum converges, it's not always possible to compute
an actual answer - you can only get only a decimal or fractional
approximation (using Taylor's theorem for the accuracy). Can you
explain to me why it's not usually possible to compute this type of
sum? Two examples of sums I'd like to find exactly are:

inf.
SUM 1/[(2n)!]
n=1
and
inf.
SUM 10^n/[(n+1)*4^(2n+1)]
n=1

By the Ratio Test, they both are absolutely convergent, so I'd like to
see if it's possible to find an exact answer. Thanks!
```

```
Date: 11/21/2000 at 00:00:52
From: Doctor Fenton
Subject: Re: If an Infinite Series Converges

Dear Dan,

You've asked a very interesting question, and in some sense, the
question is as much philosophical as mathematical. Part of the answer
to your first question, why we can't find a "closed-form" solution to
most series, in terms of "familiar" quantities, is connected with the
question, "How many irrationals do you know?"

Irrational numbers fall into two camps, the algebraic numbers that are
roots of polynomials with integer coefficients, such as sqrt(2),
sqrt(3), etc., and transcendental numbers such as pi or e. If you know
what a countable infinity is, then it will be meaningful to say that
there are only countably many finite rational combinations of
algebraic numbers, e and its powers e^2, e^3,... , pi and its powers,
and rational roots of these, but there are uncountably many real
numbers, so "most" real numbers cannot be expressed in terms of pi, e,
or algebraic numbers to begin with, no matter how the real number
arose.

If you're not familiar with countability, there's a good discussion,
"Set Theory and Orders of Infinity," here:

http://mathforum.org/dr.math/problems/klimkow.4.8.97.html

If you search our archives with a keyword like "uncountable" you'll
find even more.

So, part of the problem is that you (and I) really don't know many
irrational numbers by name. There are more real numbers than names,
for that matter.

The philosophical part of the answer is the question why you feel
more comfortable with an expression pi^2/6 than a decimal expansion.
After all, you don't really know that many decimal places for pi, and
why is writing the answer in terms of a quantity that has to be
approximated any better than just giving an expression (the series)
that will let you compute any desired number of decimal places. In
fact, the known decimal places for pi and e come from similar series
expansions. It's similar to the way most students feel more
comfortable with degrees than with radians, even though the division
of a circle into 360 parts is purely arbitrary, a 3000-year old legacy
from the Babylonians, while radians are a "natural" definition of
angles in terms of area or arclength. On the other hand, this question
does bring up the question of why you should believe in the real
numbers in the first place. Even though most people are mystified by
i, the complex unit, real numbers are much more mystifying when you
dig into them.

As for your series, those particular examples can actually be summed.
For example,

e = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + ...
and
1/e = 1 - 1 + 1/2! - 1/3! + 1/4! - 1/5! + ...
so
(e + 1/e) = 2(1 + 1/2! + 1/4! + 1/6! + ...)

which means your first series is

e + 1/e
-------  - 1
2

The second series is tougher, but from the exponents on the 10 and 4,
it resembles a geometric series. The (n+1) in the denominator can
arise from integrating x^n, so I'd begin by trying to write this as

inf.
SUM (x^n)/(n+1)
n=1

If I evaluate this series at x = 10/16, which is 10/4^2, I get your
series, except for a factor of 4 in the denominator (we'll worry about
that later.) Write

1  inf. x^(n+1)    1  inf.  x
--- SUM  ------- = --- SUM  INT t^n dt
x  n=1   (n+1)     x  n=1   0

1   x   inf.
= --- INT [SUM  t^n ] dt
x   0   n=1

1   x    t
= --- INT ----- dt
x   0   1-t

1
= --- [-x - ln(1-x)]
x

Now divide this by 4 and evaluate this result at x = 10/16, and you'll
have your second series. So, in fact, against the odds, the series you
gave just happen to have "familiar" values.

If you have further questions, please write again.

- Doctor Fenton, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Sequences, Series

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search