The Integral Test tests a series for comvergence or divergence by comparing it to an integral constructing from the series terms.
Theorem. ( Integral
Test) Suppose is a
series in which the terms are positive. Let
be the function you get by replacing k by x in
. Suppose that:
(i) f is continuous for .
(ii) f decreases for .
Then:
(a) If
converges, so does
.
(b) If diverges,
so does
.
Proof. Divide the interval up into rectangles of width 1, using the left-hand
endpoints to get the heights. The function f decreases, so the
picture looks like this:
The sum of the rectangle areas is the partial sum,
and it is clearly bigger than the area under the curve:
Therefore, if diverges, so does
, but
is the sum of the series.
Next, divide the interval up into rectangles of width
1, but use the right-hand endpoints to get the heights. The picture
looks like this:
The sum of the rectangle areas is clearly smaller than the area under the curve:
If I add to both sides, the left side becomes the
n-th partial sum:
If
converges, then the partial sums
form an increasing
sequence that is bounded above by the integral. Hence,
converges.
As an application, I'll prove the convergence criteria for p-series.
Proposition. The p-series converges if
and diverges if
.
Proof. The series has positive
terms, and
is continuous for
.
Suppose . Then
Note that implies
. Thus, in the
term
there is a positive power
of b on the bottom. Hence,
. That is,
Since the integral converges, the series converges by the Integral
Test. A similar argument shows that if , then the
series diverges.
Example. Determine whether the series converges or diverges.
is something you can integrate, so it's natural to
try the Integral Test. (And if the general term of the series is
something that would be difficult to integrate, you should try a
different test!)
For ,
. The series has positive terms.
Let . f is continuous for
. Note that
Hence, f decreases for . The hypotheses of the Integral
Test are satisfied. (It's important to check the hypothesis before
applying the test!)
Compute the improper integral:
Here's the parts table:
Since the integral converges, the series converges by the Integral
Test.
Example. Determine whether the series converges
or diverges.
The series has positive terms, and is continuous for
. Compute the
derivative:
Notice that for
. In fact, the first two
terms of the series are
and
, so the terms actually increase at the
start.
However, for
, so the terms
eventually decrease. Thus, the Integral Test applies to
(the
original series with the first term removed). Since the two series
only differ by one term, if one converges, the other converges, and
if one diverges, the other diverges.
Therefore, there's actually no harm in applying the Integral Test to the original series, even though, strictly speaking, it doesn't have decreasing terms: Eventually decreasing is good enough.
Comput the integral:
Since the integral diverges, the series diverges, by the Integral
Test.
Warning: The value of the integral in the Integral Test is not equal to the sum of the series.
The proof of the Integral Test yields the formula
You can use this to estimate the partial sums of a series to which the Integral Test applies.
Example. Estimate the sum of the first 1000 terms of
This is a divergent p-series with , so the
Integral Test applies. I have
You can compute the integrals for yourself; you'd get
Hence, .
Copyright 2019 by Bruce Ikenaga