Integral Test and p-Series
The Integral Test
Consider a series S an such that
an > 0
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:
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...
-
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.
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