Integral Test and p-Series

The Integral Test

Consider a series S an such that

an >    and     an > an+1

We can plot the points (n,an) on a graph and construct rectangles whose bases are of length 1 and whose heights are of length an.  If we can find a continuous function f(x) such that

f(n)  =  an then notice that the area of these rectangles (light blue plus purple) is an upper Reimann sum for the area under the graph of f(x).  Hence Similarly, if we investigate the lower Reimann sum we see that Since the first term is of a series has no bearing in its convergence or divergence, this proves the following theorem:

 Theorem:  The Integral Test Let f(x) be a positive continuous function that is eventually  decreasing and let f(n) = an Then converges if and only if converges.

Note the contrapositive:
 Corollary diverges if and only if diverges.

Example:

1. Consider the series: We use the Integral Test:

Let

f(x) = 1/x2 Hence by the Integral Test converges.

What does it converge to?  We use a calculator to get 1.635...

2. Consider the series We use the Integral Test.  Let  Hence by the Integral Test diverges.

P-Series Test

A special case of the integral test is when

1
an  =
np

for some p.  The theorem below discusses this.
 Theorem:  P-Series Test Consider the series If p > 1 then the series converges If 0 < p < 1 then the series diverges

Proof:

We use the integral test with the function

1
f(x)  =
xp

For p not equal to 1 Note that this limit converges if

-p + 1 < 0

or

p > 1

The limit diverges for p < 1

For p = 1 we have the harmonic series and the integral test gives: Another proof that the harmonic series diverges.