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