AP Calculus BC Unit 10: Convergence Tests for Infinite Series

Integral Test for Convergence

A big challenge with infinite series is that you can’t literally “add infinitely many terms” by hand. Convergence tests give you strategies to decide whether a series has a finite sum (converges) or grows without bound or fails to settle (diverges). The Integral Test is one of the most conceptually satisfying tests because it connects series to something you already know well from calculus: improper integrals.

What the Integral Test is

Suppose you have a series

$\sum_{n=1}^{\infty} a_n$

The Integral Test applies when the terms come from a function. Specifically, you look for a function $f$ such that

$a_n = f(n)$

and $f$ is:

• positive for $x \ge 1$
• continuous for $x \ge 1$
• decreasing for $x \ge 1$

If those conditions hold, then the Integral Test says:

$\sum_{n=1}^{\infty} a_n$

and

$\int_{1}^{\infty} f(x)\,dx$

either both converge or both diverge.

Why it matters

The Integral Test matters because many series look like “discretized versions” of areas under curves. When $f(x)$ is decreasing and positive, the rectangles of width 1 and height $f(n)$ approximate the area under $f$. If the total area under the curve from 1 to infinity is finite, then the “total rectangle area” (the series) is also finite, and vice versa.

This test is especially useful for series that resemble

$\sum \frac{1}{n^p}$

or involve logarithms, where comparisons can be tricky but integrals are manageable.

How it works (the idea)

For a positive decreasing function, the integral and the series squeeze each other. Geometrically, using left-endpoint or right-endpoint rectangles gives inequalities that relate partial sums to integrals. You don’t usually need to reproduce the proof on the AP exam, but you do need to know the conditions and be able to evaluate an improper integral.

Notation you’ll see (and what it means)

Example 1: A classic p-series via Integral Test

Determine whether

$\sum_{n=1}^{\infty} \frac{1}{n^2}$

converges.

Step 1: Match to a function. Let

$f(x) = \frac{1}{x^2}$

For $x \ge 1$, $f$ is positive, continuous, and decreasing.

Step 2: Evaluate the improper integral.

$\int_{1}^{\infty} \frac{1}{x^2}\,dx = \lim_{b\to\infty} \int_{1}^{b} x^{-2}\,dx$

Antiderivative:

$\int x^{-2}\,dx = -x^{-1} + C$

So

$\lim_{b\to\infty} \left[-\frac{1}{x}\right]_{1}^{b} = \lim_{b\to\infty} \left(-\frac{1}{b} + 1\right) = 1$

The integral converges, so the series converges.

Example 2: A series that diverges (harmonic-type)

Test

$\sum_{n=1}^{\infty} \frac{1}{n}$

Let

$f(x) = \frac{1}{x}$

Then

$\int_{1}^{\infty} \frac{1}{x}\,dx = \lim_{b\to\infty} \int_{1}^{b} \frac{1}{x}\,dx = \lim_{b\to\infty} \left[\ln x\right]_{1}^{b} = \lim_{b\to\infty} \ln b$

This diverges, so the series diverges.

What goes wrong (common pitfalls)

A frequent mistake is using the Integral Test when $f$ is not decreasing or not positive. The test is not “integral equals series”; it’s only a convergence/divergence link under specific conditions. Another common error is evaluating the improper integral incorrectly—especially forgetting the limit as the upper bound goes to infinity.

Exam Focus

• Typical question patterns:
  • “Use the Integral Test to determine convergence/divergence” (you must state conditions and compute the improper integral).
  • “Explain why the Integral Test applies” (justify positivity, continuity, decreasing behavior).
  • Occasionally: interpret convergence in terms of finite area under a curve.
• Common mistakes:
  • Forgetting to verify $f$ is decreasing on $[1,\infty)$.
  • Computing $\int_{1}^{\infty} f(x)\,dx$ but not explicitly concluding what it implies for the series.
  • Algebra errors with logs or power antiderivatives in improper integrals.

Comparison Tests

Comparison tests are your main tools when a series looks similar to a series you already understand. The big idea is: if you can trap a complicated series above or below by a simpler benchmark series, you can “inherit” convergence or divergence.

Why comparisons are so powerful

In Unit 10, you build a library of known behaviors. Two especially important benchmarks are:

• Geometric series

$\sum_{n=0}^{\infty} ar^n$

(converges if $|r|<1$)

• p-series

$\sum_{n=1}^{\infty} \frac{1}{n^p}$

(converges if $p>1$, diverges if $p\le 1$)

When you compare a new series to one of these, you often avoid more complicated tools.

Direct Comparison Test

What it is

Assume $a_n$ and $b_n$ are positive for all sufficiently large $n$ (often for all $n \ge 1$). The Direct Comparison Test uses inequalities like

$0 \le a_n \le b_n$

or

$0 \le b_n \le a_n$

to transfer convergence/divergence:

• If

$0 \le a_n \le b_n$

and

$\sum b_n$

converges, then

$\sum a_n$

also converges.

• If

$0 \le b_n \le a_n$

and

$\sum b_n$

diverges, then

$\sum a_n$

also diverges.

How to think about it

If $a_n$ is always smaller than something that adds up to a finite total, then $a_n$ can’t “accumulate” enough to diverge. Conversely, if $a_n$ is always larger than something that already blows up, then $a_n$ must blow up too.

Example 1: Convergence by comparison

Test

$\sum_{n=1}^{\infty} \frac{1}{n^2+5}$

Because $n^2+5 \ge n^2$ for $n \ge 1$,

$\frac{1}{n^2+5} \le \frac{1}{n^2}$

We know

$\sum_{n=1}^{\infty} \frac{1}{n^2}$

converges (p-series with $p=2$). Since the given terms are smaller and positive, the given series converges.

Example 2: Divergence by comparison

Test

$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}+1}$

For large $n$, this behaves like $1/\sqrt{n}$. To make a direct inequality, notice $\sqrt{n}+1 \le 2\sqrt{n}$ for $n \ge 1$, so

$\frac{1}{\sqrt{n}+1} \ge \frac{1}{2\sqrt{n}}$

Thus

$\sum_{n=1}^{\infty} \frac{1}{2\sqrt{n}} = \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$

The p-series with $p=1/2$ diverges, so the original series diverges.

Limit Comparison Test

What it is

Sometimes direct inequalities are hard to set up. The Limit Comparison Test gives a cleaner approach when two series have similar “end behavior.” For positive terms $a_n$ and $b_n$, compute

$L = \lim_{n\to\infty} \frac{a_n}{b_n}$

If $L$ is a finite positive number, meaning

$0 < L < \infty$

then

$\sum a_n$

and

$\sum b_n$

either both converge or both diverge.

Why it works (intuition)

If $a_n/b_n$ approaches a positive constant, then for large $n$ the terms $a_n$ and $b_n$ are basically constant multiples of each other. Multiplying a series by a nonzero constant doesn’t change convergence behavior, so they must “act the same.”

How to pick $b_n$

A good choice is often “the dominant term” of the denominator (or numerator) for large $n$:

• Rational functions in $n$: keep highest power of $n$.
• Roots: compare to the leading power.
• Exponentials: compare to the dominant exponential.

Example 3: Limit comparison with a p-series

Test

$\sum_{n=1}^{\infty} \frac{3n+1}{n^2+4n}$

For large $n$, the leading behavior is like

$\frac{3n}{n^2} = \frac{3}{n}$

So choose

$b_n = \frac{1}{n}$

Compute the limit:

$L = \lim_{n\to\infty} \frac{\frac{3n+1}{n^2+4n}}{\frac{1}{n}} = \lim_{n\to\infty} \frac{n(3n+1)}{n^2+4n}$

Simplify:

$\frac{n(3n+1)}{n^2+4n} = \frac{3n^2+n}{n^2+4n} = \frac{3n+1}{n+4}$

Then

$L = \lim_{n\to\infty} \frac{3n+1}{n+4} = 3$

Since $0<L<\infty$ and

$\sum_{n=1}^{\infty} \frac{1}{n}$

diverges, the original series diverges.

What goes wrong (common pitfalls)

A common misconception is thinking that $a_n < b_n$ automatically means both series do the same thing. The direction matters: you can only conclude convergence when your series is smaller than a known convergent series, and only conclude divergence when your series is larger than a known divergent series.

For Limit Comparison, students sometimes pick a bad $b_n$ (for example, something not comparable or not positive), or compute a limit of $0$ or $\infty$ and still try to conclude. If

$L=0$

or

$L=\infty$

then Limit Comparison is inconclusive (though sometimes it hints which direct comparison might work).

Exam Focus

• Typical question patterns:
  • “Determine convergence/divergence by comparison with a p-series or geometric series.”
  • “Use Limit Comparison Test; show the limit and state a conclusion.”
  • “Choose an appropriate comparison series and justify your choice.”
• Common mistakes:
  • Flipping the inequality and drawing the wrong conclusion.
  • Forgetting that comparison tests require nonnegative terms (at least eventually).
  • Using Limit Comparison when the limit is $0$ or $\infty$ and claiming a definite result.

Alternating Series Test and Error Bound

Some series don’t have all positive terms—they alternate signs. Alternating series often converge even when the corresponding positive-term series would diverge. This is the mathematical version of “cancellation”: positive and negative contributions partially undo each other.

What an alternating series is

An alternating series is typically written in the form

$\sum_{n=1}^{\infty} (-1)^{n-1} b_n$

or

$\sum_{n=1}^{\infty} (-1)^n b_n$

where $b_n \ge 0$.

The sign alternates because $(-1)^n$ switches between $1$ and $-1$.

Alternating Series Test (Leibniz Test)

What it is

The Alternating Series Test says that the series

$\sum_{n=1}^{\infty} (-1)^{n-1} b_n$

converges if both conditions hold:

1. $b_n$ is **decreasing** eventually (often you show $b_{n+1} \le b_n$ for all $n$ beyond some point).
2. $b_n \to 0$ as $n\to\infty$.

It’s crucial that the terms go to 0. If they don’t, the series cannot converge (this is the nth-term divergence idea: if $a_n$ does not go to 0, then $\sum a_n$ diverges).

Why it matters

The Alternating Series Test gives you convergence without needing integrals or comparisons, and it covers important examples like the alternating harmonic series:

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$

which converges even though

$\sum_{n=1}^{\infty} \frac{1}{n}$

diverges.

How it works (intuition)

If $b_n$ decreases to 0, then the partial sums of the alternating series “zigzag” above and below the eventual limit, with the zigzags shrinking because the terms shrink. That shrinking zigzag is what forces convergence.

Alternating Series Error Bound

What it is

One of the best features of alternating series is that you can bound the error after truncating.

Let

$S = \sum_{n=1}^{\infty} (-1)^{n-1} b_n$

and

$S_N = \sum_{n=1}^{N} (-1)^{n-1} b_n$

Then the remainder is

$R_N = S - S_N$

If the Alternating Series Test conditions hold, then

$|R_N| \le b_{N+1}$

In words: the error you make by stopping after $N$ terms is at most the next term’s magnitude.

Why it matters

AP problems often ask: “How many terms do you need to approximate the sum within a given tolerance?” For alternating series, you can answer quickly by finding $N$ such that

$b_{N+1} < \text{tolerance}$

Example 1: Convergence of an alternating series

Determine whether

$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{\sqrt{n}}$

converges.

Here

$b_n = \frac{1}{\sqrt{n}}$

Check conditions:

• $b_n$ decreases for $n\ge 1$.
• $\lim_{n\to\infty} \frac{1}{\sqrt{n}} = 0$.

So the series converges by the Alternating Series Test.

Important nuance: the corresponding positive-term series

$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$

diverges (p-series with $p=1/2$). That means the alternating series is conditionally convergent (convergent, but not absolutely convergent). While absolute vs conditional convergence is a broader topic, it’s helpful to recognize that alternating signs can “rescue” convergence.

Example 2: Using the error bound to choose $N$

Approximate

$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{n}$

to within $0.001$.

Here

$b_n = \frac{1}{n}$

We want

$|R_N| \le b_{N+1} < 0.001$

So solve

$\frac{1}{N+1} < 0.001$

This is equivalent to

$N+1 > 1000$

So you can take

$N = 1000$

That guarantees the alternating partial sum $S_{1000}$ is within $0.001$ of the true value.

What goes wrong (common pitfalls)

A common mistake is checking $b_n$ decreases but forgetting to check $b_n \to 0$. Both are required.

Another frequent error: applying the alternating series error bound to a series that is not actually alternating in sign, or where the magnitudes are not decreasing. The inequality

$|R_N| \le b_{N+1}$

depends on those conditions.

Exam Focus

• Typical question patterns:
  • “Does the alternating series converge? Justify using the Alternating Series Test.”
  • “Find the least $N$ so that the partial sum approximates the series within a given error.”
  • “Bound the error of a given partial sum.”
• Common mistakes:
  • Treating “alternating” alone as sufficient—without verifying decreasing and limit 0.
  • Using $b_N$ instead of $b_{N+1}$ in the error bound.
  • Mixing up the alternating series term $(-1)^n b_n$ with the magnitude term $b_n$ when solving inequalities.

Ratio Test for Convergence

The Ratio Test is especially effective when factorials and exponentials appear, or when terms are built from products and powers. It’s designed to measure how fast terms shrink by comparing successive terms.

What the Ratio Test is

Given a series

$\sum_{n=1}^{\infty} a_n$

(typically with nonzero terms), compute

$L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|$

Then:

• If $L < 1$, the series converges absolutely.
• If $L > 1$ (or $L = \infty$), the series diverges.
• If $L = 1$, the test is inconclusive.

Why it matters

Many series in calculus involve expressions like

$\frac{n!}{n^n}$

or

$\frac{3^n}{n!}$

where comparison tests are awkward. But ratios simplify because factorials cancel nicely:

$\frac{(n+1)!}{n!} = n+1$

The Ratio Test also connects to geometric series thinking: if the ratio of successive magnitudes approaches a number less than 1, the terms behave roughly like a geometric sequence and the series converges.

How it works (mechanism)

You compute the ratio $a_{n+1}/a_n$, simplify, take the absolute value, and then take a limit as $n\to\infty$. The limit tells you the long-run multiplicative shrinkage (or growth) from one term to the next.

Example 1: Factorials (a Ratio Test classic)

Test

$\sum_{n=1}^{\infty} \frac{n!}{3^n}$

Compute

$\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!}{3^{n+1}}\cdot\frac{3^n}{n!}$

Simplify:

$\frac{(n+1)!}{n!} = n+1$

and

$\frac{3^n}{3^{n+1}} = \frac{1}{3}$

So

$\left|\frac{a_{n+1}}{a_n}\right| = \frac{n+1}{3}$

Now take the limit:

$L = \lim_{n\to\infty} \frac{n+1}{3} = \infty$

Since $L>1$, the series diverges.

Notice what this means conceptually: terms are eventually getting larger (the ratio exceeds 1), so the series can’t possibly converge.

Example 2: A convergent series with factorial in the denominator

Test

$\sum_{n=1}^{\infty} \frac{2^n}{n!}$

Compute the ratio:

$\left|\frac{a_{n+1}}{a_n}\right| = \frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^n}$

Simplify:

$\frac{2^{n+1}}{2^n} = 2$

and

$\frac{n!}{(n+1)!} = \frac{1}{n+1}$

So

$\left|\frac{a_{n+1}}{a_n}\right| = \frac{2}{n+1}$

Take the limit:

$L = \lim_{n\to\infty} \frac{2}{n+1} = 0$

Since $L<1$, the series converges absolutely.

What goes wrong (common pitfalls)

The biggest “gotcha” is when you get $L=1$. Students sometimes think $L=1$ means diverge (confusing it with a probability-style cutoff). But for the Ratio Test, $L=1$ means you learned nothing. Many important series give $L=1$, including p-series and the harmonic series.

Another common error is algebraic: writing $a_{n+1}$ incorrectly. A careful substitution of $n+1$ everywhere in the formula for $a_n$ is essential.

Exam Focus

• Typical question patterns:
  • “Use the Ratio Test to determine convergence/divergence,” especially with factorials, exponentials, or powers.
  • “Find the limit $L$ and interpret it.”
  • Sometimes paired with alternating signs: you still use absolute values in the ratio.
• Common mistakes:
  • Concluding divergence or convergence when $L=1$ (inconclusive case).
  • Dropping absolute value bars and getting a negative limit for alternating-sign terms.
  • Miscomputing $a_{n+1}$ by only replacing some occurrences of $n$ with $n+1$.