(^610) Series
then existence of
1
aW lg(x)l dx implies existence of fag f(x) dx. Solution: The
theorem of Problem 20 implies that f If(x)l dx exists and the theorem of
a
Problem 22 then gives the required result.
(^25) Prove that if f is integrable over each finite interval and the first of the
integrals -
fo-
If(z)l dx,
fog
f(x) sin x dx,
fo,
f(x) cos x dx
exists, then the other two also exist. Solution: Since sin x and cos x are con-
tinuous, f(x) sin x and f(x) cos x are integrable over each finite interval. Since
also
If(x) sin xI < If(x)I, If(x) cos xI < lf(x)l,
the results follow from the results of the preceding problem.
12.3 Alternating series and Fourier series The following theorem
embodies the alternating series test for convergence of series.
Theorem 12.31 If the terms of a series Zuk are alternately positive and
negative and if their absolute values decrease and have the limit 0 so that
lull > IU21 > lu31 > ... , lim lu,, = 0,
then the series converges to a number s for which
(12.311)
n
s - I uk I < lun+l (n = 1,2,3, ...)
k=1
The inequality (12.311) tells us that if we use a particular partial sum
as an approximation to s, then the error will be less than the absolute
value of the first term of the series not included in the partial sum. This
information is very useful. To prove the theorem, we suppose that the
given series has the form
(12.312) a1 - a2 + as - a4 + as - a6 +
where a1 > a2 > as >. and ak -+ 0 as k ---->co. To locate the par-
tial sums s1, S2, shown in Figure 12.313, we start at the origin, go
to the right the distance a1 to reach s1j then go left the smaller distance
a2 to reach s2, then go right the still smaller distance as to reach s3j andFigure 12.313r=r
... S7 S6 S3 S1=a1