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Extra credit problems for students in math 121, focusing on improper integrals and the convergence of series. Topics include the harmonic series, alternating harmonic series, and p-series. Students are asked to show the divergence of the harmonic series, the convergence of the alternating harmonic series, and the convergence of p-series as long as |r| < 1.
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For any student planning on taking math 122, you will see topics on series. Many of these infinite series are closely related with some of the improper integrals that we look at in class.
We have touched on the idea of series a little bit in this class already. In short, series are infinite sums. As it is difficult to deal with and impossible to actually sum an infinite number of terms, we revisit the familiar idea of a limit. That is a series is an expression of the form
∞
n= 1
an = (^) Nlim→∞
N
n= 1
an
where an are numbers. In fact we have seen these types of sums in defining the definite integral to be a limit of Reimann sums ∫ (^) b a
f (x)dx = (^) Nlim→∞
N
n= 1
f (xn)b^ −^ a n
∞
n= 1
f (xn)b^ −^ a n
We use similar definitions with series as described for improperintegrals. That is a series (^) ∑∞ n= 1 an is said to converge if
Nlim→∞
N
n= 1
an
exists and is finite. The series (^) ∑∞ n= 1 an is said to diverge if
Nlim→∞
N
n= 1
an
is ±∞ or does not exist. Here we call SN = (^) ∑Nn= 1 an a partial sum as it is part of the entire infinite sum, and the above limits are called the limit of partial sums.
Problems
n
diverges. This is called the harmonic series. Hint: Interpret this series as area of rectangles, then compare it to
∫ (^) ∞ (^0) x+^11 dx. Consider the following graph.
0 2 4 6 8 10
Show that (^) ∞ ∑ n= 1
(− 1 )n+^1 n
converges. This is called the alternating harmonic series. Hint: Rewrite the series as ∞ ∑ n= 1
n
n + 1
and compare the series to ∫^1 ∞ (^) x^12 similar to the first problem.
2
ln(x)dx is divergent. Hint: Estabolish that ln(x) < x for x ≥ 2, then work from there. It may may be helpfull note that ln( 2 ) < 2 and look at the derivative of ln(x) and x for x ≥ 2.
rn
converges as long as |r| < 1. Hint: Although ∑Nn= 0 rn^ cannot be compared directly to
∫ (^) ∞ 0 rxdx, it may be compared to
∫ (^) ∞ 0 rx−^1 dx.
rn^ = (^1) −^1 r