Numerical series: convergence tests and exercises

After sequences and limits, in many Mathematical Analysis 1 (Calculus 1) courses come numerical series: between October and November, and in the winter-session written exam (eight to fourteen weeks away) there is almost always an exercise on the convergence of a series. Many universities also have a midterm in mid-December, that is nine to ten weeks from now. Calendars vary from course to course, and what counts is your own syllabus; but the time to practise is now, while the chapter is being studied, not right before the exam.

The reason is that series are a topic of choice. There are six or seven tests, and every exercise asks you to decide which one to use. Anyone who learned them as a list of statements tries at random and wastes time; anyone with a decision scheme takes two minutes. In this post you will find the scheme and four worked exercises.

What it means for a series to converge

Given a sequence $a_n$, the series $\sum a_n$ converges if the sequence of partial sums $S_N=a_1+a_2+\cdots +a_N$ converges. The value of the limit $\lim S_N$ is the sum of the series. If the limit is $\pm \infty$ the series diverges, if it does not exist the series is indeterminate (oscillating).

The first thing to remember is that for series with positive terms there is no middle way: either it converges or it diverges to $+\infty$. That is why the comparison tests work so well on them.

The reference series

Almost all exercises are solved by reducing the series to one of these two.

Geometric series. For $|q|<1$

and it diverges (or is indeterminate) for $|q|\ge 1$.

Generalised harmonic series.

converges if and only if $p>1$. So $\sum 1/n$ diverges ($p=1$), $\sum 1/\sqrt{n}$ diverges ($p=1/2$), $\sum 1/n^2$ converges. The link with improper integrals is the same: ${\int }_1^{\infty }{x}^{-p}dx$ converges precisely for $p>1$, and this is no accident (I talk about it in integrals explained properly).

A minimal example to fix the idea: with $q=1/3$ the geometric series ${\sum }_{n=0}^{\infty }(1/3{)}^n$ has sum $\frac{1}{1-1/3}=\frac{3}{2}$. The partial sums are $1,1.33,1.44,1.48,\dots$ and they approach $1.5$ without exceeding it. Watch where the index starts: if the series begins at $n=1$ instead of $n=0$ the sum changes by one term, and in the exam that is a mark lost through distraction.

The tests, one by one

Necessary condition. If $\sum a_n$ converges, then $a_n\to 0$. Read backwards it is useful: if $a_n$ does not tend to zero, the series does not converge and you are done. But the converse does not hold: in the harmonic series $a_n=1/n\to 0$ and yet the series diverges. The mistake of concluding “it converges because the term goes to zero” is by far the most frequent.

Comparison. If $0\le a_n\le b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges too. If $\sum a_n$ diverges, then $\sum b_n$ diverges too.

Asymptotic comparison. If $a_n,b_n>0$ and $a_n/b_n\to \ell$ with $0<\ell <\infty$, the two series behave the same way. It is used when $a_n$ is a complicated fraction that “looks like” a power of $n$: you keep the dominant terms.

Ratio. If $a_n>0$ and $a_{n+1}/a_n\to \ell$: with $\ell <1$ the series converges, with $\ell >1$ it diverges, with $\ell =1$ the test does not decide. Ideal with factorials and exponentials.

Root. If $\sqrt[n]{a_n}\to \ell$: the same conclusions as the ratio, with $\ell =1$ still inconclusive. Ideal when the term is an $n$-th power.

Leibniz. For an alternating-sign series $\sum (-1{)}^n{b}_n$ with $b_n\ge 0$: if $b_n$ is decreasing and $b_n\to 0$, the series converges.

Absolute convergence. If $\sum |a_n|$ converges, then $\sum a_n$ converges too. It is used for series with arbitrary sign, such as those with sine and cosine: you bound the absolute value and move on to the comparison.

Which test to use

If the ratio or root give $1$, don’t insist: move on to the asymptotic comparison.

Exercise 1: ratio test

Study

With $a_n=n/2^n$ I calculate

The limit is $1/2<1$: the series converges. (You can also calculate the sum, which is 2, but to decide convergence you don’t need it.) Note the step: $2^{n+1}$ divided by $2^n$ leaves a 2 in the denominator, and that is where people slip when in a hurry.

Before moving on to the next exercises, if you want to train the choice of test on new questions, the Analysis 1 practice test contains exercises in the same style: the correct answer is free, the step-by-step guided solution is unlocked with your first lesson.

Exercise 2: telescoping series

Study

Factorise $n^2+n=n(n+1)$ and

The partial sums cancel in a chain:

and $S_N\to 1$. The series converges and the sum is 1. Alternative without calculating the sum: $\frac{1}{n^2+n}\le \frac{1}{n^2}$, and $\sum 1/n^2$ converges ($p=2>1$), so by comparison this one converges too. In an exam choose the method that carries fewer risks: if they ask you for “the sum”, the telescoping method is compulsory.

Exercise 3: Leibniz and absolute convergence

Study, in the simple and absolute sense,

Simple convergence. It has alternating signs with $b_n=1/\sqrt{n}$. The sequence $b_n$ is decreasing and tends to zero, so by Leibniz the series converges.

Absolute convergence. The series of absolute values is $\sum 1/\sqrt{n}$, a generalised harmonic series with $p=1/2\le 1$: it diverges.

Conclusion: the series converges simply but not absolutely. It is the typical example to know because it shows that the two notions are different, and it is a classic oral question.

A case where Leibniz cannot be applied: let $b_n=1/n$ for $n$ even and $b_n=1/n^2$ for $n$ odd. Then $b_n\to 0$ but $b_n$ is not decreasing. In the series $\sum (-1{)}^n{b}_n$ the even parts are worth ${\sum }_k\frac{1}{2k}$, which diverges, while the odd ones (with a negative sign) $-\sum 1/n^2$ converge: the series diverges. So “the term goes to zero and the signs alternate” is not enough: monotonicity is a real hypothesis, not a formality.

Exercise 4: absolute convergence with the sine

Study

The sign changes irregularly, so Leibniz has no place here. I use the absolute value, with $|\sin n|\le 1$:

The series $\sum 1/n^2$ converges, so by comparison $\sum |\sin n|/n^2$ converges, and by the absolute convergence test the starting series converges too: it converges absolutely. The part you must never neglect is the order of the reasoning: first the absolute value and the comparison, then the conclusion on the original series.

Two more examples to practise the choice

Root. $\sum {\left(\frac{n}{2n+1}\right)}^n$: $\sqrt[n]{a_n}=\frac{n}{2n+1}\to \frac{1}{2}<1$, it converges.

Asymptotic comparison. $\sum \frac{n+1}{n^3+2}$: for large $n$ the term behaves like $n/n^3=1/n^2$, and $\frac{(n+1)/(n^3+2)}{1/n^2}=\frac{n^3+n^2}{n^3+2}\to 1$. Since $\sum 1/n^2$ converges, this one converges too.

The mistakes I see most

1. Concluding “it converges” because $a_n\to 0$. A necessary condition, not a sufficient one: the harmonic series $\sum 1/n$ is the counterexample.
2. Concluding from the ratio with limit 1. If $\ell =1$ the test is silent: $\sum 1/n$ and $\sum 1/n^2$ both give 1, but one diverges and the other converges.
3. Applying Leibniz without checking monotonicity, as in the example above.
4. Using the comparison with the wrong signs. Comparison with a convergent majorant is valid only if $0\le a_n\le b_n$; bounding a divergent series from above says nothing.
5. Mixing up simple and absolute convergence. The question “does it converge absolutely?” requires studying the absolute values, which are another series.

If limits of ratios slow you down, review limits explained properly and the hierarchy of infinities in sequences: limits and tests for the midterm, because almost all the tests reduce to a limit of a sequence.

How I work on this topic

With Calculus 1 students I start every exercise from the same step: looking at the general term and deciding the test before calculating, justifying the choice aloud. Then you write the conclusion with the exact statement used, because that is what the lecturer marks. I work online with a graphics tablet and a shared screen. For the overall picture there is the maths page. If you want to understand where you are losing marks, write to me: the free 30-minute introductory call comes with no commitment.