What is the difference between a sequence and a series?

A sequence is a list of numbers; a series is their sum. From a sequence (aₙ) one builds a new sequence of partial sums Sₙ = a₁ + … + aₙ, and it is the limit of that sequence which is the sum of the series. That is why the whole theory of series reduces to the limit of a sequence — the very one you already met.

What is a partial sum?

The finite sum of the first n terms, Sₙ = a₁ + a₂ + … + aₙ. Partial sums form a sequence of their own, and a series is called convergent exactly when that sequence has a finite limit. The value of that limit is then the sum of the series.

What is the necessary condition for convergence?

If a series converges, its terms must tend to zero. The condition runs one way only: when the terms do not tend to zero the series certainly diverges, but tending to zero guarantees nothing by itself. The harmonic series is the most famous counterexample.

Why does the harmonic series diverge?

Because its terms can be grouped into blocks each weighing at least one half. After the term 1/2, the block 1/3 + 1/4 exceeds 1/2, the next block from 1/5 to 1/8 does too, and so on — every doubling of the number of terms adds at least 1/2. Infinitely many such increments cannot add up to a finite limit.

When do I use the ratio test and when the root test?

The ratio test when the term contains a factorial or a product, because the ratio of consecutive terms then cancels beautifully. The root test when the whole term is raised to the n-th power, because the n-th root removes it at once. Both give the same answer whenever both apply — they differ only in how pleasant the computation is.

What does it mean that a test is inconclusive?

That the limit came out equal to 1, and nothing can be concluded from it. Both 1/n (divergent) and 1/n² (convergent) give the limit 1 in either test, so in that case another tool is needed — most often the comparison test.

What is the Leibniz test?

If the terms of an alternating series decrease in absolute value and tend to zero, the series converges. That is a very weak requirement — which is why the alternating harmonic series 1 − 1/2 + 1/3 − 1/4 + … converges (to ln 2) even though the same series without the alternating signs diverges.

What is the difference between absolute and conditional convergence?

A series converges absolutely when the series of absolute values of its terms converges, and then the series itself converges too. It converges conditionally when it converges but stops doing so once the signs are dropped — like the alternating harmonic series. The distinction matters, because rearranging the terms of a conditionally convergent series can change its sum.

Advanced level

Number series and convergence tests

A series is the limit of its sequence of partial sums — and only in a few lucky cases can that limit be computed outright. So instead of the sum, a humbler question is asked first: does the series have a sum at all. The tests answer it: comparison, ratio, root and Leibniz.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Limit of a sequence and the geometric seriesInfinitely many terms, a finite total — provided the ratio is smaller than one in absolute value. See what it means precisely for a sequence to approach a limit, how limits of polynomial quotients are computed, where the number e comes from, and why 0.(3) really is one third.

Where this is used

Real situations where you count exactly the way this lesson teaches:

A stack of blocks leaning off the table
Blocks piled one on another can hang over the edge of a table further the more of them there are: with n blocks the maximum overhang is half the n-th partial sum of the harmonic series. Four blocks give 1.04 block lengths, and to pass two lengths you need as many as 31. The overhang grows without bound, because the harmonic series diverges — it just grows desperately slowly.
How many packets to fill a sticker album
An album holds 50 different stickers and a packet contains one at random. The expected number of packets is 50 times the 50th partial sum of the harmonic series, i.e. 50·4.499 ≈ 225 packets — four and a half times the number of stickers. The last few missing slots cost more than the whole first half of the album.
How many terms to sum in a program
A program has to compute the sum of the series 1/n², which is about 1.6449. The remainder after n terms is smaller than 1/n, so an accuracy of 0.001 needs a thousand terms and an accuracy of 0.000001 already needs a million. The comparison test says the sum exists; the bound on the remainder says when the loop may stop.
Total number of decays of an isotope
A counter records decays over successive whole hours while the activity falls off exponentially: for a half-life of 6 hours the successive counts form a series with ratio e^(−ln2/6) ≈ 0.891. The ratio test gives the limit 0.891 < 1, so the series converges, and with 2 million decays in the first count the total is 2·10⁶/0.109 ≈ 1.83·10⁷.

All formulas

Partial sum
$S_n = a_1 + a_2 + \dots + a_n = \sum_{k=1}^{n} a_k$
the finite sum of the first n terms
Sum of a series
$\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} S_n$
the limit of the sequence of partial sums
Necessary condition
$\sum a_n \ \text{convergent} \implies \lim_{n \to \infty} a_n = 0$
it does not run backwards
Comparison test
$0 \leq a_n \leq b_n, \ \sum b_n \ \text{convergent} \implies \sum a_n \ \text{convergent}$
smaller than a convergent series is convergent
Ratio test
$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = g < 1 \implies \sum a_n \ \text{convergent}$
divergent for g > 1, inconclusive for g = 1
Root test
$\lim_{n \to \infty} \sqrt[n]{|a_n|} = g < 1 \implies \sum a_n \ \text{convergent}$
divergent for g > 1, inconclusive for g = 1
Leibniz test
$a_n \searrow 0 \implies \sum_{n=1}^{\infty} (-1)^{n+1} a_n \ \text{convergent}$
a sequence decreasing to zero, alternating signs
Absolute convergence
$\sum |a_n| \ \text{convergent} \implies \sum a_n \ \text{convergent}$
absolute implies plain, never the reverse

In the lesson on the limit of a sequence we summed one series — the geometric one. That worked because it has a closed formula for its partial sum, so the limit could be taken directly. Most series have no such formula, and then the question "what is the sum" has to give way to a humbler one: does a sum exist at all. This whole lesson is about the tools for that question.

A series is the limit of its partial sums

Given a sequence $(a_n)$ , build a second sequence from it — the partial sums:

S_n = a_1 + a_2 + \dots + a_n = \sum_{k=1}^{n} a_k

and define the sum of the series as the limit of that sequence:

\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} S_n

Nothing else is hiding in the definition. A series is not a new kind of object — it is a limit of a sequence, the same notion you already have from earlier lessons. When the limit is finite the series is called convergent; when it is infinite or fails to exist, divergent.

A telescoping series: the case where everything is visible

Find the sum of the series 1/(1·2) + 1/(2·3) + 1/(3·4) + …

The partial sums climb towards the line S = 1 and never reach it. That line is the sum of the series — the limit of the sequence of dots, not one of them.

A telescope, though, is a rare luxury. Usually there is no formula for $S_n$ and convergence has to be settled without knowing the sum.

The necessary condition and its trap

If a series converges, $S_n$ and $S_{n-1}$ share the same limit, and their difference is $a_n$ . Hence, immediately:

\sum a_n \ \text{convergent} \implies \lim_{n \to \infty} a_n = 0

The condition is necessary but not sufficient — and that distinction is the most important thing in this lesson. It is useful in its contrapositive form: if the terms do not tend to zero, the series diverges. The series $\sum \tfrac{n}{n+1}$ has terms tending to $1$ , so it is out at once, with no test needed.

In the other direction nothing follows from it, and the most famous witness is the harmonic series.

Show that the harmonic series 1 + 1/2 + 1/3 + 1/4 + … diverges.

Partial sums of the harmonic series at n = 1, 2, 4, 8, …, 64, with the line 1 + k/2 beneath them. The line rises without end and stays below the dots, so the dots have nowhere to settle either.

The sum of $64$ terms is still only $4.74$ — the divergence here is exceptionally slow, and that makes no difference at all. Divergence is a question of the bound, not of the pace.

The comparison test

The simplest tool: measure your series against one you already know everything about. For series with non-negative terms:

0 \leq a_n \leq b_n \quad \Longrightarrow \quad \begin{cases} \sum b_n \ \text{convergent} \implies \sum a_n \ \text{convergent} \\ \sum a_n \ \text{divergent} \implies \sum b_n \ \text{divergent} \end{cases}

The yardsticks are two series you already have: the geometric one, $\sum q^n$ , convergent exactly for $|q| < 1$ , and the harmonic one, $\sum \tfrac{1}{n}$ , divergent. A third is the family $\sum \tfrac{1}{n^p}$ , convergent for $p > 1$ .

Decide whether the series Σ 1/(n² + 3) converges.

The ratio test

Instead of comparing against a yardstick, one can look at how fast the terms shrink relative to each other:

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

$g$	conclusion
$g < 1$	series convergent
$g > 1$	series divergent
$g = 1$	test inconclusive

The idea is simple: when $g < 1$ , the terms eventually shrink faster than a geometric sequence with ratio $g$ , and such a series converges. So the test reduces an unknown series to a known one.

Use the ratio test on the series Σ 2ⁿ/n!.

The case $g = 1$ really does mean nothing. Both $\sum \tfrac{1}{n}$ (divergent) and $\sum \tfrac{1}{n^2}$ (convergent) give $g = 1$ . When that happens, go back to comparison.

The root test

A twin tool, convenient when the whole term sits under an n-th power:

g = \lim_{n \to \infty} \sqrt[n]{|a_n|}

The conclusions are exactly those of the ratio test: $g < 1$ convergent, $g > 1$ divergent, $g = 1$ inconclusive.

Decide whether the series Σ ((n + 1)/(3n))ⁿ converges.

Alternating series and the Leibniz test

When the signs of the terms flip at every step, the partial sums jump above and below the limit in turn — and that alone helps a great deal.

a_n \searrow 0 \quad \Longrightarrow \quad \sum_{n=1}^{\infty} (-1)^{n+1} a_n \ \text{converges}

There are two requirements and both are modest: the sequence $(a_n)$ has to be decreasing and to tend to zero.

Decide whether the alternating harmonic series 1 − 1/2 + 1/3 − 1/4 + … converges.

There is a practical bonus too: in a series meeting Leibniz's conditions, the truncation error is smaller than the first discarded term. Summing a hundred terms of the alternating harmonic series is off by less than $\tfrac{1}{101}$ .

Absolute and conditional convergence

A series is called absolutely convergent when $\sum |a_n|$ converges. Then

\sum |a_n| \ \text{convergent} \implies \sum a_n \ \text{convergent}

The implication does not run backwards, and the alternating harmonic series is the example: it converges, and once the signs are dropped it diverges. Such a series is called conditionally convergent.

series	$\sum a_n$	$\sum \lvert a_n \rvert$	kind of convergence
$\sum \frac{(-1)^{n+1}}{n^2}$	convergent	convergent	absolute
$\sum \frac{(-1)^{n+1}}{n}$	convergent	divergent	conditional
$\sum \frac{1}{n}$	divergent	divergent	none

The difference is not a formality. In an absolutely convergent series the terms may be rearranged and the sum will not change. In a conditionally convergent one they may not: by rearranging the alternating harmonic series one can reach any sum decided in advance.

Choosing a test

Do the terms tend to zero? If not — done, the series diverges.
Is there a factorial or a power with exponent $n$ ? Factorials and products → ratio test; the whole term under an $n$ -th power → root test.
Do the signs alternate? Then Leibniz, and check separately whether the convergence is absolute.
Does the term look like $\tfrac{1}{n^p}$ ? Then the comparison test and the family of yardsticks.
Did you get $g = 1$ ? The test is inconclusive — go back to point 4.

Exercises

The set mixes four computations: the partial sum of a telescoping series, the sum of such a series in the limit, and the value of the limit computed by the ratio test and by the root test. The answer is always a number or a fraction, e.g. 3/4.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

Partial sum of the series: aₖ = 1/(k(k + 1)); S₄

Notice what the set does not ask: the verdict of convergent-or-divergent. That verdict is one sentence drawn from the number you have just computed — compare $g$ with 1 and the table above says the rest. The divergence proof for the harmonic series stays in the text too: its answer is an argument, not a result.

Common mistakes

Assuming that terms tending to zero make the series convergent — that is the necessary condition run backwards, and the harmonic series refutes it.
Confusing a sequence with a series — the sequence $\tfrac{1}{n}$ converges to zero while the series $\sum \tfrac{1}{n}$ diverges. Two different questions about the same numbers.
Treating $g = 1$ as an answer — it is the one case in which both tests stay silent; another tool is required.
Comparing series with negative terms — the comparison test needs non-negative terms; with mixed signs, examine $\sum |a_n|$ first.
Applying the Leibniz test without checking monotonicity — tending to zero is not enough; the sequence has to decrease as well.
Rearranging a conditionally convergent series — the sum can then come out as anything at all, so that is no innocent manipulation.

Formula card

Topic: Number series

Partial sum
$S_n = a_1 + a_2 + \dots + a_n = \sum_{k=1}^{n} a_k$
the finite sum of the first n terms
Sum of a series
$\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} S_n$
the limit of the sequence of partial sums
Necessary condition
$\sum a_n \ \text{convergent} \implies \lim_{n \to \infty} a_n = 0$
it does not run backwards
Comparison test
$0 \leq a_n \leq b_n, \ \sum b_n \ \text{convergent} \implies \sum a_n \ \text{convergent}$
smaller than a convergent series is convergent
Ratio test
$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = g < 1 \implies \sum a_n \ \text{convergent}$
divergent for g > 1, inconclusive for g = 1
Root test
$\lim_{n \to \infty} \sqrt[n]{|a_n|} = g < 1 \implies \sum a_n \ \text{convergent}$
divergent for g > 1, inconclusive for g = 1
Leibniz test
$a_n \searrow 0 \implies \sum_{n=1}^{\infty} (-1)^{n+1} a_n \ \text{convergent}$
a sequence decreasing to zero, alternating signs
Absolute convergence
$\sum |a_n| \ \text{convergent} \implies \sum a_n \ \text{convergent}$
absolute implies plain, never the reverse

Partial sums of the telescoping series 1/(k(k+1)). Each one lies closer to the line S = 1 and none of them reaches it, because Sₙ = 1 − 1/(n+1).

Partial sums of the harmonic series at n = 2^k. The line 1 + k/2 is a lower bound that nothing stops — which is why the series diverges even though its terms tend to zero.

Frequently asked questions

A luminous parabola and a straight line crossing above a faint coordinate grid, the vertex flaring like a star, against a cosmic nebula in violet and magenta

Functions and analysis

Why a function's graph can never double back

The whole definition of a function comes down to one word: uniqueness. One argument, one value — and not a single one more. That seemingly minor condition decides why a circle is not the graph of a function, why a line and a parabola describe so much of the world, and why four datasets with identical summary statistics can look nothing alike.

12 min