Difficulty: Intermediate | Prerequisites: Limits at infinity, sigma notation, sequences
This material covers the fundamental types of infinite series you will meet in Calculus 1 (or early Calculus 2): the Divergence Test, telescoping series, and geometric series. It sits after sequences in the syllabus and before the heavier convergence tests (integral, comparison, ratio, root). If you are not comfortable evaluating limits, especially limits involving fractions and factored expressions, sort that out first. Everything here depends on it.
An infinite series converges only if its sequence of partial sums approaches a finite limit. Telescoping series collapse when you decompose terms with partial fractions. Geometric series have a clean closed-form sum whenever the common ratio has absolute value less than 1. The Divergence Test is your first checkpoint: if the terms do not tend to zero, stop, the series diverges.
Infinite series
The sum of infinitely many terms, written as the sum from n = 1 (or 0) to infinity of a_n. In simple terms, you are adding up a list of numbers that never ends.
Partial sum (S_k)
The sum of the first k terms of a series. Think of it as a running total: S_k = a_1 + a_2 + ... + a_k.
Convergence (of a series)
A series converges if the sequence of its partial sums approaches a finite number as k tends to infinity. In simple terms, the running total settles on a specific value.
Divergence (of a series)
A series diverges if the partial sums do not approach a finite limit. The running total either blows up, oscillates, or otherwise refuses to settle.
Divergence Test (nth-term test)
If lim (n to infinity) a_n is not equal to 0, the series diverges. Think of it as a necessary condition: if the individual terms are not shrinking to zero, there is no chance the sum converges. Note: the converse is false. Terms going to zero does not guarantee convergence.
Telescoping series
A series whose partial sums collapse because most terms cancel in pairs. After partial-fraction decomposition, you typically get a difference of two simple fractions, and nearly everything cancels when you add them up.
Geometric series
A series of the form a + ar + ar^2 + ar^3 + ..., where r is the common ratio. It converges to a / (1 - r) when |r| < 1 and diverges when |r| >= 1.
Common ratio (r)
The fixed multiplier between consecutive terms in a geometric series: r = a_(n+1) / a_n. In simple terms, each term is the previous term times r.
Consider the series: sum from n = 1 to infinity of cos(nπ).
The values of cos(nπ) cycle through -1, 1, -1, 1, ... The terms do not approach zero. The partial sums oscillate between -1 and 0 and never settle on a value.
By the Divergence Test, this series does not converge.
Applying the Ratio Test here: the ratio |a_(n+1) / a_n| = |cos((n+1)π) / cos(nπ)| = 1 for every n. The Ratio Test is inconclusive when the limit equals 1, so it gives no answer here. You fall back on the Divergence Test, which settles it immediately.
Key takeaway: the Divergence Test is the simplest and fastest test. Always check it first. If lim a_n is not 0, you are done. If lim a_n = 0, the test is silent and you need a different tool.
Consider the series: sum from n = 1 to infinity of a_n, where a_n = 5 / (n^2 + 3n).
Step 1: Factor the denominator.
n^2 + 3n = n(n + 3)
Step 2: Partial-fraction decomposition.
5 / [n(n + 3)] = A/n + B/(n + 3)
Multiplying through: 5 = A(n + 3) + Bn. Setting n = 0 gives A = 5/3. Setting n = -3 gives B = -5/3.
So a_n = (5/3)(1/n - 1/(n + 3)).
Step 3: Write out the partial sums and watch them telescope.
S_k = (5/3) [ (1/1 - 1/4) + (1/2 - 1/5) + (1/3 - 1/6) + (1/4 - 1/7) + ... ]
Most terms cancel. What survives at the left end is 1/1 + 1/2 + 1/3 = 11/6. The tail terms 1/(k+1) + 1/(k+2) + 1/(k+3) vanish as k tends to infinity.
Step 4: Take the limit.
lim S_k = (5/3)(1 + 1/2 + 1/3) = (5/3)(11/6) = 55/18.
The worksheet uses the decomposition a_n = 1/n - 1/(n+3) (without the leading 5/3 factor). In that version, the surviving terms are 1 + 1/2 + 1/3 and the tail vanishes, giving lim S_k = 11/6. The essential technique is the same either way:
The worksheet uses a slightly different decomposition. The essential technique is the same:
Factor the denominator
Decompose into partial fractions
Write out enough terms of S_k to see the cancellation pattern
Identify the surviving terms and take the limit
Why it works: each term a_n is a difference of two pieces. In the running sum, the positive piece of one term cancels the negative piece of a later term, collapsing the whole sum to a handful of boundary terms.
A geometric series has the form: sum from n = 0 to infinity of a * r^n = a / (1 - r), provided |r| < 1.
Worked example from the worksheet:
Given a recursive sequence where a_(n+1) = a_n / 3 and a_0 = 10.
This means each term is 1/3 of the previous term. The common ratio is r = 1/3.
The series is: 10 + 10/3 + 10/9 + 10/27 + ...
Since |r| = 1/3 < 1, it converges.
Sum = a / (1 - r) = 10 / (1 - 1/3) = 10 / (2/3) = 15.
The formula unpacked:
The partial sum of a geometric series is S_k = a(1 - r^(k+1)) / (1 - r). As k tends to infinity, r^(k+1) tends to 0 (because |r| < 1), and you are left with a / (1 - r).
When it diverges:
If |r| >= 1, the terms do not shrink, and the partial sums grow without bound (or oscillate). The Divergence Test catches this: lim a * r^n is not 0 when |r| >= 1.
Students often think that if a_n tends to 0, the series must converge. It does not. The harmonic series (sum of 1/n) is the classic counterexample: terms go to zero, but the sum diverges. The Divergence Test only works in one direction.
Students frequently forget to check the Divergence Test first and jump straight into heavier tests. Always do the quick check: does a_n tend to 0? If not, you already have your answer.
When telescoping, students sometimes fail to write out enough terms to spot which ones survive. Write at least five or six terms. The cancellation pattern is not always obvious from two or three.
With geometric series, students mix up the first term and the common ratio. The formula a / (1 - r) requires a = the first term actually written, and r = the multiplier. If the sum starts at n = 1 instead of n = 0, the first term is ar, not a.
⚠️ The Divergence Test is almost always the first thing to check on an exam problem. If you skip it and dive into a harder test, you waste time.
⚠️ Telescoping series are a favourite exam question because the technique has clear, testable steps: factor, decompose, write partial sums, cancel, take the limit. Expect to be asked for the exact sum, not just "converges or diverges."
⚠️ Geometric series appear in nearly every exam, often disguised. Rewriting a series into the form a * r^n is a skill worth drilling. Common disguises include (-1)^n * (1/2)^n (which is just r = -1/2) and sums that start at n = 3 instead of n = 0.
⚠️ Know the geometric series formula cold: sum = a / (1 - r) for |r| < 1. Exams will not give it to you.
True or false: if lim (n to infinity) a_n = 0, then the series sum of a_n converges.
Fill in the blank: a geometric series sum from n = 0 to infinity of ar^n converges when ______.
True or false: the series sum of cos(nπ) converges.
Fill in the blank: the sum of a convergent geometric series with first term a and ratio r is ______.
True or false: in a telescoping series, you need partial fractions to split the general term before you can see the cancellation.
(Answers: 1. False, the harmonic series is a counterexample. 2. |r| < 1. 3. False, the terms alternate between -1 and 1 and do not approach 0. 4. a / (1 - r). 5. Generally true for rational terms, though some telescoping series arise from other algebraic manipulations.)
Q: Find the sum of the series: sum from n = 0 to infinity of 6 * (1/4)^n.
A: This is a geometric series with a = 6 and r = 1/4. Sum = 6 / (1 - 1/4) = 6 / (3/4) = 8.
Q: Determine whether the series sum from n = 1 to infinity of n / (n + 1) converges or diverges.
A: lim (n to infinity) of n/(n+1) = 1, which is not 0. By the Divergence Test, the series diverges.
Q: Use partial fractions to find the sum of: sum from n = 1 to infinity of 1 / [n(n + 1)].
A: 1/[n(n+1)] = 1/n - 1/(n+1). The partial sum S_k = (1 - 1/2) + (1/2 - 1/3) + ... + (1/k - 1/(k+1)) = 1 - 1/(k+1). As k tends to infinity, S_k tends to 1. The sum is 1.
Q: A geometric series has a_0 = 10 and each term is 1/3 of the previous. What is the sum?
A: a = 10, r = 1/3. Sum = 10 / (1 - 1/3) = 10 / (2/3) = 15.
Q: Does the series sum from n = 1 to infinity of (-1)^n converge?
A: The terms alternate between -1 and 1 and do not approach 0. By the Divergence Test, the series diverges. The partial sums oscillate between -1 and 0.
This material connects directly to the convergence tests covered in the next unit (comparison, integral, ratio, root tests), which you need when a series is neither geometric nor telescoping. The partial-fraction technique used for telescoping series also appears in integral calculus when integrating rational functions.
Geometric series return in Taylor and Maclaurin series: the function 1/(1 - x) is literally the sum of x^n, and many Taylor expansions build on that identity.
Infinite series, partial sums, sequence of partial sums, telescoping series, telescoping sum, partial fraction decomposition, geometric series, geometric sum formula, common ratio, convergent series, divergent series, Divergence Test, nth-term test, nth term test for divergence, harmonic series, MATH 140, Calculus 1, Penn State, series convergence, sum to infinity