Lecture 32
Wednesday, March 27, 2013
Math 103 (Section 208 - Pooya Ronagh)
As an application of what we have learned in chapter 9, we can add one test to our arsenal
of convergence/divergence tests for series:
This is applied to series ∑∞
k=N ak that are expressed using a function. So for this test to
be useful you need a function y = f (x) such that ak is the value of f at the integer point
x = k:
ak = f (k).
In other words our series is ∑∞
k=N f (k). Then we have a
Theorem 1. Let y = f (x) be a function defined on the interval [N, +∞). We also assume
that f (x) is non-negative (i.e. for all points x in [N, +∞), we have f (x) ≥ 0). Then
+∞
1. If the improper integral ∫N
verges. And,
+∞
2. If the improper integral ∫N
f (x)dx converges, then the series ∑∞
k=N f (k) also con-
f (x)dx diverges, then the series ∑∞
k=N f (k) also diverges.
1
1
Example 1. Is ∑∞
n=2 n ln(n) convergent? Answer: No! Since n ln n are values of f (x) =
∞
1
1
x ln(x) at integer points, we only need to check convergence of the improper integral ∫2 x ln(x) dx:
∞
∫
2
M
M
1
1
dx = lim ∫
dx = lim ln(ln(x))∣ = ∞
M →∞ 2
M →∞
2
x ln(x)
x ln(x)
hence divergence of the improper integral, ergo divergence of the series.
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