Sequences, series and convergence tests notes — Unit 5

A Sequence is just an ordered list of numbers separated by commas: 1, 1/2, 1/4, 1/8...
A Series is the SUM of those numbers: 1 + 1/2 + 1/4 + 1/8 + ...

⇒The Big Question: Convergence

If you add up an infinite amount of positive numbers, won't the sum just equal infinity? Not always. If the numbers shrink to zero fast enough, the infinite sum will equal a perfectly finite number. We call this 'Convergence'. If it explodes to infinity, we call it 'Divergence'.

Before testing a series, you must check the most basic rule. If the terms of the series don't eventually shrink down to exactly zero, there is absolutely no hope for the series to converge.

Test the series: Σ [ n / (n + 1) ]
uₙ = n / (n + 1)
lim(n→∞) [ n / (n + 1) ] = lim(n→∞) [ 1 / (1 + 1/n) ] = 1 / (1 + 0) = 1.
Since the limit is 1 (not 0), the series diverges.

To test a complex series, we usually compare it to a simple series whose behavior we already know. You must memorize the rules for these two benchmark series.

If you have a complex series uₙ, find a simpler series vₙ to compare it to.

1. 1.Example: Test Σ [ 1 / (n² + 1) ]
2. 2.1. We know n² + 1 > n²
3. 3.2. Therefore, 1 / (n² + 1) < 1 / n²
4. 4.3. Let uₙ = 1 / (n² + 1) and vₙ = 1 / n²
5. 5.4. Since vₙ is a p-series with p=2 (which is > 1), it converges.
6. 6.5. Since uₙ is smaller than a convergent series, uₙ also converges.

The Direct Comparison test requires you to prove an inequality (uₙ < vₙ), which can be algebraically annoying. The Limit Comparison test skips the inequality entirely.

The secret to this test is knowing how to choose the auxiliary series vₙ. You simply extract the highest power of 'n' from the numerator, and divide it by the highest power of 'n' from the denominator.

Test the series: Σ [ √(n³ + 1) / (n³ + 3n² + 2) ]

1. 1.Step 1: Construct the auxiliary series vₙ.
2. 2.Highest power in numerator = √(n³) = n^(3/2)
3. 3.Highest power in denominator = n³
4. 4.vₙ = n^(3/2) / n³ = 1 / n^(3 - 3/2) = 1 / n^(3/2)

Step 2: Evaluate lim (uₙ / vₙ).
lim [ √(n³+1) / (n³+3n²+2) ] / [ 1 / n^(3/2) ]
= lim [ √(n³+1) * n^(3/2) ] / [ n³+3n²+2 ]
Factor out n³ from top and bottom to cancel:
= lim [ n^(3/2) * √(1 + 1/n³) * n^(3/2) ] / [ n³(1 + 3/n + 2/n³) ]
= lim [ n³ * √(1 + 1/n³) ] / [ n³(1 + 3/n + 2/n³) ]
= lim [ √(1 + 0) ] / [ 1 + 0 + 0 ]
= 1

If a series contains factorials (n!) or exponentials (xⁿ), the Comparison tests fail because you can't easily extract a power of 'n'. You MUST use D'Alembert's Ratio Test.

Test the series: Σ [ n! / nⁿ ]

1. 1.Step 1: Write uₙ and uₙ₊₁.
2. 2.uₙ = n! / nⁿ
3. 3.uₙ₊₁ = (n+1)! / (n+1)ⁿ⁺¹

Step 2: Calculate uₙ₊₁ / uₙ
= [ (n+1)! / (n+1)ⁿ⁺¹ ] * [ nⁿ / n! ]
Expand factorials: (n+1)! = (n+1) * n!
= [ (n+1)n! / (n+1)(n+1)ⁿ ] * [ nⁿ / n! ]
Cancel n! and (n+1):
= nⁿ / (n+1)ⁿ
= [ n / (n+1) ]ⁿ
= [ 1 / (1 + 1/n) ]ⁿ
= 1 / (1 + 1/n)ⁿ

1. 1.Step 3: Take the limit as n→∞.
2. 2.Recall the standard limit: lim(n→∞) (1 + 1/n)ⁿ = 'e'.
3. 3.L = 1 / e.

In almost every 10-mark exam question, the series will contain a variable 'x'. You will apply the ratio test and find L = x. You will write: 'Converges for x<1, diverges for x>1'. But what about when x=1?

When x=1, the ratio test completely fails (L=1). You must substitute x=1 back into your original formula for (uₙ₊₁ / uₙ), and apply a stronger test. The two backup tests are Raabe's Test and the Logarithmic Test.

Raabe's Test is slightly more sensitive than the Ratio test. It can detect convergence even when the ratio is exactly 1.

Assume we tested a series, put x=1, and found that (uₙ / uₙ₊₁) = (3n² + 4n) / (3n² + 2).

1. 1.We must apply Raabe's Test because lim(n→∞) of this fraction is exactly 1.
2. 2.Step 1: Subtract 1 from the fraction.
3. 3.(uₙ / uₙ₊₁) - 1 = [ (3n² + 4n) / (3n² + 2) ] - 1
4. 4. = [ 3n² + 4n - (3n² + 2) ] / (3n² + 2)
5. 5. = (4n - 2) / (3n² + 2)

Step 2: Multiply by 'n'.
n * [ (uₙ / uₙ₊₁) - 1 ] = n * [ (4n - 2) / (3n² + 2) ]
                       = (4n² - 2n) / (3n² + 2)

1. 1.Step 3: Take the limit as n→∞.
2. 2.Divide top and bottom by n².
3. 3.= lim [ 4 - 2/n ] / [ 3 + 2/n² ]
4. 4.= 4 / 3

If the ENTIRE nth term is raised to the power of 'n', don't bother with the Ratio test. It will create massive algebraic headaches. Use Cauchy's Root Test.

Test: Σ [ (n+1) / 2n ]ⁿ
uₙ = [ (n+1) / 2n ]ⁿ
(uₙ)^(1/n) = [ (n+1) / 2n ]
lim(n→∞) = lim [ (1 + 1/n) / 2 ] = 1/2.
Since 1/2 < 1, it Converges.

Everything so far assumed all terms were positive. What if the signs alternate? (+, -, +, -, +). This is called an Alternating Series: Σ (-1)ⁿ⁻¹ uₙ.

If you have an alternating series, there are two 'levels' of convergence.

Alternating Harmonic Series: 1 - 1/2 + 1/3 - 1/4...
Passes Leibnitz? Yes (1/n shrinks to 0). So it converges.
Take absolute values: 1 + 1/2 + 1/3 + 1/4...
This is the Harmonic p-series (p=1). It diverges!
Therefore, 1 - 1/2 + 1/3... is Conditionally Convergent.

In the exam, they won't tell you which test to use. Use this exact mental flowchart to decide in 5 seconds.

A 10-mark question almost always asks to 'Discuss the convergence of the series for all positive values of x'. Let's look at the exact structure of the answer you must provide to get full marks.

1. 1.Phase 1: Apply Ratio Test.
2. 2.Calculate L = lim(uₙ₊₁ / uₙ). You will get an answer containing 'x', for example, L = x/3.
3. 3.Phase 2: State the standard conditions.
4. 4.Write clearly: 'By Ratio test, series converges if x/3 < 1 (x < 3) and diverges if x/3 > 1 (x > 3). Test fails at x = 3.'
5. 5.Phase 3: Resolve the failure.
6. 6.Substitute x = 3 into the original (uₙ / uₙ₊₁) fraction. Apply Raabe's Test.
7. 7.Phase 4: The Final Conclusion.
8. 8.Combine everything. 'Converges for x ≤ 3, diverges for x > 3.'

Discuss the convergence of: x + (x²)/2 + (x³)/3 + (x⁴)/4 + ...

Step 1: Write the general term uₙ.
uₙ = xⁿ / n
uₙ₊₁ = xⁿ⁺¹ / (n+1)

1. 1.Step 2: Apply Ratio Test.
2. 2.uₙ₊₁ / uₙ = [ xⁿ⁺¹ / (n+1) ] * [ n / xⁿ ]
3. 3. = x * [ n / (n+1) ]
4. 4. = x * [ 1 / (1 + 1/n) ]
5. 5.
6. 6.Limit as n→∞ = x * (1/1) = x.
7. 7.So, L = x.
8. 8.Converges if x < 1, Diverges if x > 1. Fails at x = 1.

Now we must test the case where x = 1.

Step 3: Substitute x=1 into the original series.
The series becomes: 1 + 1/2 + 1/3 + 1/4 + ...
This is exactly the Harmonic p-series: Σ (1/n).
Here p = 1.

1. 1.Step 4: Use the p-series rule.
2. 2.Since p = 1 (which is not > 1), the p-series test tells us it DIVERGES.
3. 3.
4. 4.Step 5: Write the final combined conclusion.
5. 5.The series converges strictly for x < 1.
6. 6.It diverges for x ≥ 1.

A series like Σ Cₙxⁿ is called a Power Series. The 'Radius of Convergence' (R) is the maximum value of 'x' for which the series converges. If it converges for x < 3, then R = 3.

If R = 0, the series only converges at exactly x=0. If R = ∞, the series converges for literally any number you plug in (like the Maclaurin series for eˣ).

⇒Unit 5 Mastery

• Do you remember the p-series rule (converges if p > 1, diverges if p ≤ 1)?
• Can you correctly extract the highest powers of n for the Limit Comparison Test?
• Do you know the Ratio Test conditions (L<1 converges)?
• Do you remember to FLIP the fraction upside down for Raabe's Test?
• Can you state the two rules of Leibnitz's test for alternating series?
• Can you perfectly reproduce the 4-step flowchart for choosing a test?