The Art and Craft of Problem Solving

9.4 POWER SERIES AND EULERIAN MATHEMATICS 343

provided that this limit exists.
Note that the an (x) are all defined for all real x, so the same is true for each Sn(x).
In fact, we can explicitly compute Sn(x), using the formula for a geometric series (see
page 158); we have

x^2 x^2 x^2

Sn(x) =

1 +x 2

+

(1 +x (^2) ) 2

• ... +
  (1 +x2)n
  x^2 x^2
  (1 +x2) (1 +x2)n+1
  1
  1 - --
  (^1) +x 2
  Multiply numerator and denominator by 1 + x^2 and we get
  x^2
  x^2 �=-
  S
  ()_ (1+x2)n
  n x -
  X^2
  This formula is not defined if x = O. As long as x =I-0, we can simplify further to get
  1
  Sn(x) = 1 -
  (1 +x2)n·
  Fix a real number x =I-O. As n -+ 00, the second term above will vanish, no matter
  what x is. Therefore S(x) = 1 for all nonzerox. But if x = 0, then each an (0) = 0; this
  forces Sn(O) = 0 and we conclude that S(O) = O. •
  That wasn't too bad, but something disturbing happened. Each an(x) is continuous
  (in fact, differentiable), yet the infinite sum of these functions is discontinuous. This
  example warns us that infinite series of functions cannot be treated like finite series.
  There are plenty of other "pathologies:" for example, a function f(x) is defined to be
  the infinite sum of fi(X), yet f'(x) is not equal to the sum of the ff
  (
  x).l^0 The basic rea­
  son behind these troubles is the fact that properties like continuity and differentiation
  involve taking limits, as does finding the sum of a series. It is not alway s the case that
  a "limit of a limit" is unchanged when you interchange the order.
  Luckily, there is one key property that prevents most of these pathologies: uni­
  fo rm convergence, which is defined in the same spirit as uniform continuity (see
  page 324 ). We say that the sequence of functions Un (x) ) converges uniformly to

f(x) if the "N response" to the "e challenge" is independent of x. We shall not discuss

many details here (see [36] for a clear and concise treatment, and [29] for a fresh and
intuitive discussion) because the punch line is, "Don't worry." Here's why.

• If fn(x) -+ f(x) uniformly, and the fn are continuous, then f(x) will also be
  continuous.

IOSee Chapter 7 of [36] for a nice discussion of these issues. (Our Example 9. 4. 1 was adapted from this
chapter.)