Convergence and Divergence Tests for Series and Justifications

AP Calculus BC: Infinite Series Convergence & Divergence TestsIntroduction & Master Summary Table Before diving into the algebra, it is crucial to understand what we are doing: we are determining whether an infinite sum of numbers eventually settles on a finite value (converges) or grows to infinity/fails to settle (diverges). The table below is your cheat sheet. Knowing when to use a test is just as important as knowing how to use it.Test NameSeries Form / When to UseQualifications / ConditionsConclusion (Proves Conv, Div, or Both)1. nth Term TestAny series ∑an​Find limn→∞​an​Divergence ONLY. (If liman​=0, it diverges).2. Geometric Series∑a⋅rnIdentify common ratio rBoth. Conv if ∣r∣<1; Div if ∣r∣≥1.3. p-Series∑np1​Identify exponent pBoth. Conv if p>1; Div if p≤1.4. Alternating Series (AST)∑(−1)nbn​Terms must alternate signsConvergence ONLY. (Proves Conv if bn+1​≤bn​ and limbn​=0).5. Ratio TestSeries with n! or cnFind L=limn→∞​​an​an+1​​​ConditionL=limn→∞​​an​an+1​​​<1L=limn→∞​​an​an+1​​​>1 or âˆžL=limn→∞​​an​an+1​​​=1​ConclusionAbsolutely ConvergentDivergentInconclusive​​6. Direct Comparison (DCT)Algebraic fractions / trig0≤an​≤bn​ (positive terms)Both. Big conv ⟹ Small conv. Small div ⟹ Big div.7. Limit Comparison (LCT)Messy algebraic fractionsan​,bn​>0 (positive terms)Both. If limbn​an​​=L (0<L<∞), both do the same.8. Integral TestEasy to integrate formsf(x) is positive, continuous, decreasingBoth. ∫1∞​f(x)dx and ∑an​ share the same fate.1. The nth Term Test for DivergenceThe Concept: If the individual terms of a sequence do not shrink to exactly zero, there is absolutely no way their infinite sum can be finite. The Trap: If limn→∞​an​=0, you know NOTHING. The test is inconclusive. Never say a series converges by the nth term test!Example 1.1 (MCQ Style)Determine the convergence or divergence of ∑n=1∞​4n2+13n2−5​.Justification: We take the limit of the terms: limn→∞​4n2+13n2−5​=43​. Conclusion: Since limn→∞​an​=43​=0, the series diverges by the nth Term Test for Divergence.Example 1.2 (FRQ Style) Does the series ∑n=1∞​nsin(n1​) converge or diverge? Justify your answer.Justification: Let's find the limit of the nth term. limn→∞​nsin(n1​). This is a 0⋅∞ indeterminate form. Let x=n1​. As n→∞,x→0. The limit becomes limx→0​xsin(x)​=1.Conclusion: Because limn→∞​an​=1=0, the series diverges by the nth Term Test for Divergence.2. Geometric Series TestThe Concept: A series of the form ∑n=0∞​a⋅rn. If ∣r∣<1, it converges to the sum S=1−rFirst Term​.If ∣r∣≥1, it diverges.Example 2.1 (Finding the Sum) Determine if ∑n=1∞​5n3n+1​ converges or diverges. If it converges, find its sum.Justification: Rewrite the general term to easily see a and r: 5n3n+1​=5n3⋅3n​=3(53​)n. This is a geometric series with common ratio r=53​. Conclusion: Since ∣r∣=53​<1, the geometric series converges. The Sum: The first term (plug in n=1) is 3(3/5)1=59​. The sum is S=1−3/59/5​=2/59/5​=29​.Example 2.2 (Tricky r Value) Does the series ∑n=2∞​πne2n​ converge or diverge?Justification: Rewrite as ∑(πe2​)n. The common ratio is r=πe2​. Since e≈2.718, e2≈7.38. Therefore, r≈3.147.38​>1.Conclusion: Because ∣r∣=πe2​≥1, the series diverges by the Geometric Series Test.3. p-Series TestThe Concept: A series of the form ∑n=1∞​np1​, where p is a positive constant. Converges if p>1.Diverges if p≤1. (Note: if p=1, this is ∑n1​, known as the Harmonic Series, which strictly diverges).Example 3.1 Determine the convergence of ∑n=1∞​3n4​5​.Justification: Pull out the constant: 5∑n=1∞​n4/31​. This is a p-series with p=34​.Conclusion: Since p=34​>1, the series converges by the p-Series Test. (Multiplying by a constant 5 does not affect convergence).Example 3.2 Does ∑n=1∞​n−0.99 converge or diverge?Justification: Rewrite the series with a positive exponent in the denominator: ∑n=1∞​n0.991​. This is a p-series with p=0.99.Conclusion: Since p=0.99≤1, the series diverges by the p-Series Test.4. Alternating Series Test (AST)The Concept: For a series ∑n=1∞​(−1)nbn​ or ∑n=1∞​(−1)n+1bn​ (where bn​>0), the series converges if TWO conditions are met:bn+1​≤bn​ for all n (the positive terms are strictly decreasing).limn→∞​bn​=0.Example 4.1 (Standard FRQ) Does ∑n=1∞​n2+1(−1)n+1​ converge or diverge? Justify.Justification: We are evaluating the alternating series where bn​=n2+11​. We must verify the two conditions of AST:Limit to zero: limn→∞​n2+11​=0.Decreasing: (n+1)2+11​<n2+11​ for all n≥1, so bn+1​<bn​.Conclusion: Since bn​ is decreasing and approaches 0, the series converges by the Alternating Series Test.Example 4.2 (The AST Trap) Does ∑n=1∞​2n+5(−1)n⋅3n​ converge or diverge?Justification: Let bn​=2n+53n​. We check the AST limit condition: limn→∞​2n+53n​=23​. Because the limit of the non-alternating part is NOT zero, the overall limit limn→∞​an​ does not exist (it oscillates between ≈1.5 and ≈−1.5).Conclusion: The series diverges by the nth Term Test for Divergence. (Note: DO NOT say "diverges by AST". AST can only prove convergence!)5. Ratio TestThe Concept: Evaluate L=limn→∞​​an​an+1​​​. Best used for: Exponentials (like 3n) and Factorials (like n!).If L<1, the series converges (absolutely). If L>1, it diverges. If L=1, the test is inconclusive (use another test).Example 5.1 (Factorial and Exponential Combo) Determine if ∑n=1∞​nnn!2n​ converges or diverges. (Wait, nn is a bit complex for a standard ratio, let's look at a more standard AP example!) Let's evaluate ∑n=1∞​n!5n​.Justification: Apply the Ratio Test. L=limn→∞​​(n+1)!5n+1​⋅5nn!​​ =limn→∞​​(n+1)⋅n!5⋅5n​⋅5nn!​​=limn→∞​​n+15​​=0.Conclusion: Since L=0<1, the series converges by the Ratio Test.Example 5.2 (Algebraic & Exponential) Evaluate ∑n=1∞​22nn23n​.Justification: Note that 22n=4n. We apply the Ratio Test: L=limn→∞​​4n+1(n+1)23n+1​⋅n23n4n​​ =limn→∞​​n2(n+1)2​⋅43​​=1⋅43​=43​.Conclusion: Since L=43​<1, the series converges by the Ratio Test.6. Direct Comparison Test (DCT)The Concept: Compare your series an​ to a known series bn​ (usually a p-series or geometric series). Both series must have strictly positive terms.If the "Big" series converges, the "Small" series converges (an​≤bn​).If the "Small" series diverges, the "Big" series diverges (an​≥bn​).Example 6.1 (Bounding Trig Functions) Does ∑n=1∞​n32+cos(n)​ converge or diverge?Justification: We know that −1≤cos(n)≤1. Therefore, the numerator is always positive and bounded: 1≤2+cos(n)≤3. Let an​=n32+cos(n)​ and compare it to bn​=n33​. We state the inequality: 0<n32+cos(n)​≤n33​ for all n≥1. The series ∑n33​ is a convergent p-series (p=3>1).Conclusion: Because 0<an​≤bn​ and ∑bn​ converges, the original series converges by the Direct Comparison Test.Example 6.2 (Denominator Comparison) Determine convergence of ∑n=1∞​n​−0.51​.Justification: Let an​=n​−0.51​. We want to compare this to bn​=n​1​. Because we are subtracting 0.5 from the denominator, the denominator is smaller, which makes the overall fraction larger. Therefore, n​−0.51​>n​1​>0. We know ∑n​1​ diverges because it is a p-series with p=1/2≤1.Conclusion: Because an​>bn​>0 and the smaller series ∑bn​ diverges, the original series diverges by the Direct Comparison Test.7. Limit Comparison Test (LCT)The Concept: When DCT inequalities go the "wrong way" or the algebra is messy polynomial fractions, use LCT. Let an​>0 and bn​>0. Calculate L=limn→∞​bn​an​​. If 0<L<∞ (a positive, finite constant), then both series share the exact same fate (both converge or both diverge).Example 7.1 (Polynomial Fractions) Determine the convergence of ∑n=1∞​n4+2n+13n2−1​.Justification: Both numerator and denominator eventually act like their leading terms. We choose our comparison series bn​=n4n2​=n21​. Note that an​,bn​>0 for n≥1. Let's evaluate the limit: L=limn→∞​bn​an​​=limn→∞​(n4+2n+13n2−1​⋅1n2​)=limn→∞​n4+2n+13n4−n2​=3. Since L=3 (which is finite and >0), LCT applies. We know ∑n21​ is a convergent p-series (p=2>1).Conclusion: Because 0<L<∞ and the comparison series converges, the original series converges by the Limit Comparison Test.Example 7.2 (Radicals) Does ∑n=1∞​n4+3​5n​ converge or diverge?Justification: The dominant term in the top is n. The dominant term in the bottom is n4​=n2. We compare an​ to bn​=n2n​=n1​. Both an​,bn​>0. Find the limit: limn→∞​bn​an​​=limn→∞​n4+3​5n​⋅1n​=limn→∞​n4+3​5n2​=limn→∞​n21+3/n4​5n2​=5. Since L=5 (finite, >0), LCT applies. ∑n1​ is the harmonic series (or p-series with p=1), which diverges.Conclusion: By the Limit Comparison Test, the original series diverges.8. Integral TestThe Concept: If an​=f(n), evaluate the improper integral ∫1∞​f(x)dx. If the integral converges to a number, the series converges. If the integral diverges, the series diverges. The AP Catch: You MUST explicitly state that the function f(x) is positive, continuous, and decreasing for x≥1 to earn the condition point on an FRQ.Example 8.1 (U-Substitution) Determine the convergence of ∑n=1∞​n2+1n​.Justification: Let f(x)=x2+1x​. For x≥1, f(x) is clearly positive and continuous (denominator is never 0). It is decreasing because f′(x)=(x2+1)2(x2+1)(1)−x(2x)​=(x2+1)21−x2​, which is ≤0 for x≥1. We evaluate ∫1∞​x2+1x​dx=limb→∞​∫1b​x2+1x​dx. Let u=x2+1⟹du=2xdx⟹21​du=xdx. =limb→∞​[21​ln∣x2+1∣]1b​=limb→∞​21​(ln(b2+1)−ln(2))=∞.Conclusion: Since the improper integral diverges, the series diverges by the Integral Test.Example 8.2 (lnx Integrals) Does ∑n=2∞​nlnn1​ converge or diverge?Justification: Let f(x)=xlnx1​. For x≥2, f(x) is positive, continuous, and decreasing. ∫2∞​xlnx1​dx=limb→∞​∫2b​xlnx1​dx. Let u=lnx⟹du=x1​dx. =limb→∞​[ln∣lnx∣]2b​=limb→∞​(ln(lnb)−ln(ln2))=∞.Conclusion: Since the improper integral diverges, the series diverges by the Integral Test.Absolute vs. Conditional ConvergenceA series ∑an​ is:Absolutely Convergent if ∑∣an​∣ converges. (If the absolute value converges, the original series converges).Conditionally Convergent if ∑∣an​∣ diverges, but the original alternating series ∑an​ converges (usually by AST).Divergent if both the original and absolute value series diverge.Example A: Determine if ∑n=1∞​n3(−1)n​ converges absolutely, conditionally, or diverges.Justification: First, test the absolute value: ∑​n3(−1)n​​=∑n31​. This is a p-series with p=3>1, which converges.Conclusion: Because the absolute value of the series converges, the series converges absolutely.Example B: Determine if ∑n=1∞​n​(−1)n+1​ converges absolutely, conditionally, or diverges.Justification: First, test the absolute value: ∑​n​(−1)n+1​​=∑n1/21​.This is a p-series with p=1/2≤1, so the absolute value diverges.Next, test the original alternating series using AST:bn​=n​1​. limn→∞​n​1​=0. n+1​1​<n​1​, so it is decreasing. The original series converges by AST.Conclusion: Because the absolute value diverges but the alternating version converges, the series converges conditionally.Example C: Determine if ∑n=2∞​lnn(−1)nn​ converges absolutely, conditionally, or diverges.Justification: Check the limit of the general term: limn→∞​lnnn​. Applying L'Hopital's Rule, we get limn→∞​1/n1​=limn→∞​n=∞.Since the limit does not equal 0, the terms do not go to 0.Conclusion: The series diverges by the nth term test.Example D: Determine if ∑n=1∞​n!(−1)n3n​ converges absolutely, conditionally, or diverges.Justification: Because of the n! and exponential 3n, we jump straight to the Ratio Test, which tests for absolute convergence directly! L=limn→∞​​(n+1)!(−1)n+13n+1​⋅(−1)n3nn!​​=limn→∞​n+13​=0.Conclusion: Since L=0<1, the series converges absolutely by the Ratio Test.Strategy: How to Attack Problems & Choose the Right TestWhen you are given a random series on the AP test without being told which test to use, follow this mental flowchart:nth Term Check: Do the terms clearly not go to zero in your head? If so, state it diverges by the nth Term Test.Special Forms: Is it easily recognizable as a p-series (1/np) or Geometric Series (a⋅rn)?Alternating: Does it have an (−1)n? Use the Alternating Series Test (AST).Factorials (n!) or Exponentials (cn mixed with algebra): Immediately use the Ratio Test.Messy Algebraic Fractions (Polynomials/Roots): Use Limit Comparison Test (LCT) comparing to the dominant terms.Functions easily integrated (like lnn/n): Use the Integral Test.Mixed Practice (5 Examples)Problem 1: Evaluate the convergence of ∑n=1∞​n7+1​2n2+3n​Attack Strategy: Messy algebraic fraction. Let's use LCT. The top acts like n2, bottom acts like n7​=n3.5. n2/n3.5=1/n1.5.Full Justification: Compare to bn​=n1.51​. Both an​,bn​>0. L=limn→∞​n7+1​2n2+3n​⋅1n1.5​=limn→∞​n7+1​2n3.5+3n2.5​=limn→∞​n7​2n3.5​=2. Because 0<L<∞ and the comparison series ∑n1.51​ converges (p=1.5>1), the given series converges by the Limit Comparison Test.Problem 2: Evaluate the convergence of ∑n=1∞​100nn!​Attack Strategy: Factorial and exponential means Ratio Test, immediately.Full Justification: Apply the Ratio Test: L=limn→∞​​100n+1(n+1)!​⋅n!100n​​=limn→∞​​100n+1​​=∞. Because L=∞>1, the series diverges by the Ratio Test.Problem 3: Evaluate the convergence of ∑n=2∞​ln(n)(−1)n​Attack Strategy: It's alternating, so we use AST. Full Justification: Let bn​=ln(n)1​. limn→∞​lnn1​=0.Since ln(n+1)>ln(n), we know ln(n+1)1​<ln(n)1​, so the terms are strictly decreasing.Because both conditions are met, the series converges by the Alternating Series Test.Problem 4: Evaluate the convergence of ∑n=1∞​ne−n2Attack Strategy: This looks exactly like a u-substitution problem. Integral Test.Full Justification: Let f(x)=xe−x2. For x≥1, f(x) is positive, continuous, and decreasing (f′(x)=e−x2(1−2x2)<0 for x≥1).Evaluate ∫1∞​xe−x2dx=limb→∞​∫1b​xe−x2dx.Let u=−x2⟹du=−2xdx⟹−21​du=xdx.limb→∞​[−21​e−x2]1b​=limb→∞​(2eb2−1​−2e−1​)=0+2e1​.Because the improper integral converges to 2e1​, the series converges by the Integral Test.Problem 5: Evaluate the convergence of ∑n=1∞​(1+n1​)nAttack Strategy: Check the nth term limit first. This is a very famous limit definition in Calculus.Full Justification: Let's take the limit as n→∞ of the sequence: limn→∞​(1+n1​)n. By definition, this limit evaluates to Euler's number, e≈2.718.Because limn→∞​an​=e=0, the terms do not approach zero. Therefore, the series diverges by the nth Term Test for Divergence.