Welcome to Part 19 of our 21 Part series: Calculus Revisited. We are in the middle of working with series. Today's session: Tests of Convergence.
In a calculus course, you are often asked if certain series converge - that is, does the series have a sum. You are not necessarily asked to find the sum itself, just to say if the series has one.
Hint: This is good to know, the p-series.
For a series
∞
∑ 1/n^p
n = 1
this series converges when p > 1 and diverges when p ≤ 1.
Now for the tests:
Comparison Test:
Give the infinite series ∑ a(n), find another comparable infinite series ∑ b(n). The series used for comparison dominates the series in question. Or:
b_n ≥ a_n for each n an integer
In addition, all terms both series, ∑ a(n) and ∑ b(n), are positive.
If ∑ b(n) converges, so does ∑ a(n).
However, if ∑ a(n) diverges, so does ∑ b(n).
There is a variant of the comparison test call the limit test. That is if
lim ( a_n / b_n) = L as n → ∞,
If ∑ b(n) converges and L < ∞, ∑ a(n) converges.
If ∑ b(n) diverges and L > 0, ∑ a(n) diverges.
Ratio Test:
For a given infinite series ∑ a(n), if
lim | a_n+1 / a_n | = L as n → ∞, and
L < 1,
then ∑ a(n) is absolutely convergent. If L = 1, the test is inconclusive. If L > 1, the series diverges.
Note: A series is absolutely convergent if
∑ |a(n)| = | a(1) | + | a(2) | + | a(3) | + ....
is convergent.
If ∑ a(n) converges but ∑ | a(n) | does not, then the series is said to be conditionally convergent.
Root Test:
For a given infinite series ∑ a(n), if
lim | a_n |^(1/n) = L as n → ∞, and
L < 1,
then ∑ a(n) is absolutely convergent. If L = 1, the test is inconclusive. If L > 1, the series diverges.
Alternating Series Test:
For the existing series:
∑ (-1)^n * a(n),
if the sequence {a_n} (ignoring the (-1)^n) is a strictly decreasing sequence of positive numbers (that is a_n+1 < a_n for all n), and
lim a_n = 0 as n → ∞,
the the alternating series is (at least) conditionally convergent.
Problems
1. Does this series converge?
∞
∑ n^2 / (2n - 1)!
n = 1
Use the ratio test. Then:
a_n = n^2 / (2n - 1)!
a_n+1 = (n + 1)^2 / (2(n + 1) - 1)! = (n + 1)^2 / (2n + 1)!
a_n+1 / a_n
= (n + 1)^2 / (2n + 1)! * (2n - 1)! / n^2
= (n + 1)^2 / (n^2 * (2n + 1)(2n))
= (n^2 + 2n + 1) / (4n^4 + 2n^3)
As n → ∞, a_n+1 / a_n → 0 < 1.
By the ratio test, this series converges.
Sometimes reasoning will be necessary.
2. Does this series converge?
∞
∑ 1 / (n * n^(1/n))
n = 1
Note that:
1 / (n * n^(1/n))
= 1 / (n^1 * n^(1/n))
= 1 / (n^(1 + 1/n))
Compare the series to the harmonic series:
∞
∑ 1/n
n = 1
where 1/n ≥ 1/(n^(1 + 1/n))
By Limit test:
lim ( 1/(n^(1 + 1/n)) / (1/n) ) as n → ∞
= lim ( n / n^(1+ 1/n) ) as n → ∞
As n grows large, 1 + 1/n → 1
Hence, n / n^(1 + 1/n) → n / n → 1
Since ∑ (1/n) diverges, so does ∑ 1/ (n * n^(1/n))
3. Does this series converge?
∞
∑ 1/(4n - 3)
n = 1
Use the limit test. Compare it to the harmonic series, which is known to diverge, and:
1/n ≥ 1/(4n - 3)
lim ( (1/(4n - 3)) / (1/n) ) as → ∞
= lim ( n / (4n - 3) ) as → ∞
= 1/4 > 0
Hence, the series in question diverges.
4. Does this series converge?
1 - 1/√3 + 1/√5 - 1/√7 + .... + (-1)^n/√(2n+1) + ....
Use the Alternating Series test.
Let a_n = 1/√(2n+1)
First a_n strictly decreases. Second, a_n → 0 as n → ∞.
By alternating series, the series conditionally convergent.
5. Does this series converge?
1/(2 ln 2) - 1/(3 ln 3) + 1/(4 ln 4) - 1/(5 ln 5) + ... + (-1)^n/(n ln n) + ....
Use the Alternating Series Test
Let a_n = 1/(n ln n),
a_n strictly decreases and a_n → 0 as n → ∞.
By alternating series, the series is conditionally convergent.
Next time we work with Taylor Series.
See you soon!
Eddie
This blog is property of Edward Shore. © 2012