Adding Infinitely Many Numbers: A Friendly Guide to Series with Positive Terms
Can you add up infinitely many numbers and get a finite answer? Sometimes yes, sometimes no. This post explains how to tell the difference, one test at a time, with worked examples for each.
1. The big idea: partial sums
Imagine walking towards a wall 2 meters away. First you walk 1 meter, then half a meter, then a quarter, then an eighth, and so on. You take infinitely many steps, yet you never pass the wall. The total distance is exactly 2 meters:
That is the whole spirit of infinite series. We cannot literally perform infinitely many additions, so instead we add finitely many terms and watch what happens as we add more and more.
Definition
A (numerical) series is an expression
\[ a_1 + a_2 + a_3 + \cdots + a_n + \cdots = \sum_{n=1}^{\infty} a_n. \tag{1} \]
Its n-th partial sum is \( S_n = a_1 + a_2 + \cdots + a_n \).
The series converges if the finite limit
\[ \lim_{n\to\infty} S_n = S \]
exists; the number \(S\) is called the sum of the series. Otherwise the series diverges.
So a series is really just a sequence in disguise: the sequence of partial sums \(S_1, S_2, S_3, \dots\). Convergence of the series means convergence of that sequence.
Example 1 · Geometric series
For \( \sum_{n=0}^{\infty} r^n \) we have \( S_n = 1 + r + \cdots + r^{n-1} = \dfrac{1 - r^n}{1 - r} \) (for \(r \ne 1\)).
If \(|r| < 1\), then \(r^n \to 0\), so
\[ \sum_{n=0}^{\infty} r^n = \frac{1}{1-r}. \]
If \(|r| \ge 1\), the partial sums do not settle down and the series diverges. With \(r = \tfrac12\) we get \(2\), the wall from our story.
Example 2 · A telescoping series
Since \( \dfrac{1}{n(n+1)} = \dfrac1n - \dfrac{1}{n+1} \), almost everything cancels:
\[ S_n = \Big(1 - \tfrac12\Big) + \Big(\tfrac12 - \tfrac13\Big) + \cdots + \Big(\tfrac1n - \tfrac1{n+1}\Big) = 1 - \frac{1}{n+1} \longrightarrow 1. \]
Hence \( \sum_{n=1}^{\infty} \dfrac{1}{n(n+1)} = 1 \).
In this post we focus on series with positive (non-negative) terms, \(a_n \ge 0\). For those, the partial sums can only grow: \(S_{n+1} = S_n + a_{n+1} \ge S_n\). A growing sequence has only two options: it either stays bounded (and then converges) or it shoots off to \(+\infty\). That simple fact is the engine behind every test below.
Key fact
A series with non-negative terms converges if and only if its partial sums are bounded.
2. The Cauchy criterion and the n-th term test
Computing \(S_n\) in closed form, as in the examples above, is rarely possible. Cauchy's criterion lets us decide convergence without knowing the sum.
Cauchy criterion
The series (1) converges if and only if for every \(\varepsilon > 0\) there is a natural number \(N = N(\varepsilon)\) such that for all \(n > N\) and all \(p > 0\)
\[ |S_{n+p} - S_n| = |a_{n+1} + a_{n+2} + \cdots + a_{n+p}| < \varepsilon. \]
In plain words: a series converges exactly when its "tails" become negligible. Far enough out, any block of consecutive terms, however long, adds up to something tiny.
Taking \(p = 1\) gives \(|a_{n+1}| < \varepsilon\), which leads to the most basic test of all:
Necessary condition (n-th term test)
If \( \sum a_n \) converges, then \( \lim_{n\to\infty} a_n = 0 \).
Equivalently: if \(a_n \not\to 0\), the series diverges.
The most common mistake
The converse is false. Terms going to zero does not guarantee convergence. The classic counterexample is the harmonic series
\[ 1 + \frac12 + \frac13 + \frac14 + \cdots. \]
Group its terms:
\[ 1 + \frac12 + \underbrace{\Big(\frac13 + \frac14\Big)}_{> \frac12} + \underbrace{\Big(\frac15 + \cdots + \frac18\Big)}_{> \frac12} + \underbrace{\Big(\frac19 + \cdots + \frac1{16}\Big)}_{> \frac12} + \cdots \]
Each group exceeds \(\tfrac12\), and there are infinitely many groups, so the sum is infinite, even though \(\tfrac1n \to 0\).
3. Comparison tests
The idea is the same one you use when shopping: if your basket costs less than a friend's basket and your friend can afford theirs, you can afford yours. Let a second series be given:
Comparison test
Suppose that for all \(n > n_0\)
\[ 0 \le a_n \le b_n. \]
Then:
if the "bigger" series (2) converges, the "smaller" series (1) converges too;
if the "smaller" series (1) diverges, the "bigger" series (2) diverges too.
Note the phrase "for \(n > n_0\)". The first few terms never affect whether a series converges (they only affect the value of the sum), so the inequality only needs to hold eventually.
3.2 Limit comparison
Finding an exact inequality can be fiddly. Often it is easier to check that two series "behave alike" for large \(n\).
Limit comparison test
Let \(a_n \ge 0\), \(b_n > 0\) and
\[ \lim_{n\to\infty} \frac{a_n}{b_n} = k. \]
If \(0 < k < \infty\), the two series converge or diverge together.
If \(k = 0\) and \(\sum b_n\) converges, then \(\sum a_n\) converges.
If \(k = \infty\) and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.
3.3 The p-series: your best comparison partner
To use comparison tests you need a library of series whose behaviour you already know. The single most useful one is
\[ \sum_{n=1}^{\infty} \frac{1}{n^p}, \quad \text{which converges for } p > 1 \text{ and diverges for } p \le 1. \]
(We prove this in Section 8 with the integral test.) So \(\sum \frac1{n^2}\) converges, \(\sum \frac1{\sqrt n}\) diverges, and \(p = 1\) is the harmonic series, sitting right on the borderline and diverging. Together with the geometric series, the p-series covers most comparisons you will ever need.
Example 3 · Direct comparison
\( \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + n + 1} \). Since \(n^2 + n + 1 > n^2\), we have \( \dfrac{1}{n^2+n+1} < \dfrac1{n^2} \). The p-series with \(p = 2\) converges, so ours converges too.
Example 4 · Limit comparison
\( \displaystyle\sum_{n=1}^{\infty} \frac{2n + 3}{n^3 + 1} \). For large \(n\), the numerator behaves like \(2n\) and the denominator like \(n^3\), so the term behaves like \( \frac{2}{n^2} \). Take \(b_n = \frac1{n^2}\):
\[ \lim_{n\to\infty} \frac{(2n+3)/(n^3+1)}{1/n^2} = \lim_{n\to\infty} \frac{2n^3 + 3n^2}{n^3 + 1} = 2. \]
Since \(0 < 2 < \infty\) and \(\sum \frac1{n^2}\) converges, the series converges.
Tip: for rational expressions in \(n\), keep only the highest powers in the numerator and denominator. The difference of degrees tells you the answer immediately: the series converges exactly when (degree of denominator) − (degree of numerator) > 1.
4. D'Alembert's ratio test
Comparison tests need a partner series. The next tests compare the series with a geometric series automatically, by looking only at the series itself. The question the ratio test asks is: by what factor does each term shrink compared to the previous one?
D'Alembert's test
Let \(a_n > 0\).
If \( \displaystyle\limsup_{n\to\infty} \frac{a_{n+1}}{a_n} = q < 1 \), the series converges.
If \( \displaystyle\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = q > 1 \), the series diverges.
If \(q = 1\), the test says nothing.
Why it works: if eventually \( a_{n+1} \le r\, a_n \) for some \(r < 1\), then \( a_{n_0+k} \le a_{n_0} r^k \). The terms are dominated by a convergent geometric series. If the ratio is above 1, the terms eventually grow, so they cannot tend to zero.
(The lim sup is the largest limit point of the sequence of ratios. When the ordinary limit exists, it is just that limit, which is the usual case in exercises.)
Example 5
\( \displaystyle\sum_{n=1}^{\infty} \frac{2^n}{n!} \):
\[ \frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^n} = \frac{2}{n+1} \longrightarrow 0 < 1. \]
The series converges. (In fact, its sum is \(e^2 - 1\).)
Example 6
\( \displaystyle\sum_{n=1}^{\infty} \frac{n!}{n^n} \):
\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!} = \Big(\frac{n}{n+1}\Big)^{n} = \frac{1}{\big(1 + \frac1n\big)^n} \longrightarrow \frac1e < 1. \]
The series converges.
When the ratio test fails
For every p-series, \( \frac{a_{n+1}}{a_n} = \big(\frac{n}{n+1}\big)^p \to 1 \), whether it converges (\(p = 2\)) or diverges (\(p = 1\)). So whenever the terms look like powers of \(n\), don't waste time on the ratio test. It shines with factorials and exponentials.
5. Cauchy's root test
The root test asks a similar question in a different way: does \(a_n\) behave like \(q^n\) for some \(q\)? To find out, take the n-th root.
Cauchy's root test
Let \(a_n \ge 0\) and
\[ \limsup_{n\to\infty} \sqrt[n]{a_n} = q. \]
If \(q < 1\) the series converges; if \(q > 1\) it diverges; if \(q = 1\) the test is inconclusive.
Example 7
\( \displaystyle\sum_{n=1}^{\infty} \Big(\frac{n}{2n+1}\Big)^{n} \):
\[ \sqrt[n]{a_n} = \frac{n}{2n+1} \longrightarrow \frac12 < 1. \]
The series converges. Whenever the whole term is raised to the power \(n\), the root test is the natural choice.
Good to know: the root test is slightly stronger than the ratio test. Whenever the ratio test gives an answer, the root test gives the same answer, but not always the other way round. In practice, choose whichever makes the algebra easier.
6. Raabe's test
What do we do when the ratio test gives \(q = 1\)? We look more closely at how fast the ratio approaches 1. Raabe's test is a magnifying glass for exactly this borderline case.
Raabe's test
Let \(a_n > 0\) and
\[ \lim_{n\to\infty} n\left(\frac{a_n}{a_{n+1}} - 1\right) = p. \]
If \(p > 1\) the series converges; if \(p < 1\) it diverges. (For \(p = 1\) the test is inconclusive.)
Note that the fraction is flipped compared to the ratio test: here it is \( \frac{a_n}{a_{n+1}} \). The intuition comes from the p-series: for \(a_n = \frac1{n^p}\) one gets \( n\big(\frac{a_n}{a_{n+1}} - 1\big) \to p \). Raabe's test effectively compares your series with \( \sum \frac1{n^p} \).
Ratio test: \( \frac{a_{n+1}}{a_n} = \frac{2n+1}{2n+2} \to 1 \). No decision. Now apply Raabe:
\[ n\left(\frac{a_n}{a_{n+1}} - 1\right) = n\left(\frac{2n+2}{2n+1} - 1\right) = \frac{n}{2n+1} \longrightarrow \frac12 < 1. \]
The series diverges.
7. Gauss's test
Gauss's test combines the ratio test and Raabe's test into a single statement, and it even settles the case Raabe leaves open.
Gauss's test
Let \(a_n > 0\) and suppose the ratio can be written as
\[ \frac{a_n}{a_{n+1}} = \lambda + \frac{\mu}{n} + \frac{\theta_n}{n^{1+\varepsilon}}, \]
where \(\theta_n\) is bounded and \(\varepsilon > 0\). Then:
the series converges if \(\lambda > 1\), or if \(\lambda = 1\) and \(\mu > 1\);
the series diverges if \(\lambda < 1\), or if \(\lambda = 1\) and \(\mu \le 1\).
How to read it: \(\lambda\) is the "ratio test part" (reciprocal of \(q\)), \(\mu\) is the "Raabe part", and the last term is a small error that does not matter. The great bonus is the case \(\lambda = 1,\ \mu = 1\): Gauss tells us the series diverges, as the harmonic series does.
In practice, you get the expansion by writing \( \frac{a_n}{a_{n+1}} \) as a function of \( \frac1n \) and expanding (for example, with the binomial series or Taylor series) up to the \( \frac1{n^2} \) term.
Example 9
For \(a_n = \frac1{n^p}\):
\[ \frac{a_n}{a_{n+1}} = \Big(1 + \frac1n\Big)^p = 1 + \frac{p}{n} + \frac{\theta_n}{n^2}. \]
So \(\lambda = 1\), \(\mu = p\), \(\varepsilon = 1\). Gauss gives convergence for \(p > 1\) and divergence for \(p \le 1\), exactly the p-series rule, including the borderline \(p = 1\).
Example 10 · Revisiting Example 8, with squares
Consider \( \displaystyle\sum_{n=1}^{\infty} \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \). Here
\[ \frac{a_n}{a_{n+1}} = \left(\frac{2n+2}{2n+1}\right)^2 = \left(1 + \frac{1}{2n+1}\right)^2 = 1 + \frac{1}{n} + \frac{\theta_n}{n^2}. \]
We get \(\lambda = 1\), \(\mu = 1\). Raabe gives \(p = 1\) and stays silent, but Gauss says: diverges.
8. The integral test
The last tool connects series with something you already know well: improper integrals. Picture the terms \(a_n = f(n)\) as rectangles of width 1 and height \(f(n)\) standing under (or over) the graph of \(f\). If the area under the curve is finite, the rectangles cannot add up to infinity, and the reverse also holds.
Cauchy's integral test
Let \(f(x)\), \(x \ge 1\), be a non-negative, non-increasing (decreasing) function. Then the series
\[ \sum_{n=1}^{\infty} f(n) \]
and the improper integral
\[ \int_1^{\infty} f(x)\,dx \]
either both converge or both diverge.
The precise reason: since \(f\) is decreasing, on each interval \([k, k+1]\) we have \( f(k+1) \le f(x) \le f(k) \), and integrating gives
So the partial sums are bounded exactly when the integrals are bounded.
Watch out
The function must be decreasing. This is essential: it's what lets us trap the sum between areas. Also remember that the test says the integral and the sum converge together; it does not say they are equal.
Example 11 · Proof of the p-series rule
Take \(f(x) = \frac1{x^p}\) with \(p > 0\) (decreasing and positive). For \(p \ne 1\):
\[ \int_1^{\infty} \frac{dx}{x^p} = \lim_{b\to\infty} \frac{b^{1-p} - 1}{1-p} = \begin{cases} \dfrac{1}{p-1}, & p > 1,\\[4pt] \infty, & p < 1, \end{cases} \]
and for \(p = 1\): \( \int_1^{\infty} \frac{dx}{x} = \lim_{b\to\infty} \ln b = \infty \). (For \(p \le 0\) the terms do not tend to zero.) This proves that \( \sum \frac1{n^p} \) converges exactly when \(p > 1\).
Example 12 · A series that diverges very slowly
\( \displaystyle\sum_{n=2}^{\infty} \frac{1}{n \ln n} \). With \(f(x) = \frac{1}{x \ln x}\) (positive and decreasing for \(x \ge 2\)) and the substitution \(t = \ln x\):
\[ \int_2^{\infty} \frac{dx}{x \ln x} = \lim_{b\to\infty} \big(\ln\ln b - \ln\ln 2\big) = \infty. \]
The series diverges, even though it diverges more slowly than the harmonic series. (The partial sum after a billion terms is only about 3.)
9. Which test should I use?
With this many tests, the real skill is choosing the right one quickly. Here is a practical checklist:
Do the terms go to zero? If \(a_n \not\to 0\), you are done: the series diverges.
Is it a known series? Geometric (\(\sum r^n\)), p-series (\(\sum \frac1{n^p}\)), or telescoping.
Is the term a rational function of \(n\), or does it contain roots of polynomials? Use (limit) comparison with a p-series.
Does it contain factorials, or \(c^n\)? Try the ratio test.
Is the entire term raised to the n-th power? Try the root test.
Did the ratio test give \(q = 1\)? Move to Raabe's test, and if that also gives 1, to Gauss's test.
Is \(a_n = f(n)\) for an easy-to-integrate decreasing function (for example, involving \(\ln n\))? Use the integral test.
Try each one before opening the solution. More solved problems from this topic are in Series.
Problem 1: \( \displaystyle\sum_{n=1}^{\infty} \frac{n}{3n + 5} \)
\( a_n \to \frac13 \ne 0 \), so the series diverges by the n-th term test.
Problem 2: \( \displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3 + 2}} \)
The term behaves like \( \frac{1}{n^{3/2}} \). Limit comparison with \(b_n = \frac{1}{n^{3/2}}\) gives \(k = 1\). Since \(p = \frac32 > 1\), the series converges.
Problem 3: \( \displaystyle\sum_{n=1}^{\infty} \frac{n^2}{3^n} \)
Ratio test: \( \frac{a_{n+1}}{a_n} = \frac{(n+1)^2}{3n^2} \to \frac13 < 1 \). The series converges.
Problem 4: \( \displaystyle\sum_{n=1}^{\infty} \Big(1 - \frac1n\Big)^{n^2} \)
Root test: \( \sqrt[n]{a_n} = \big(1 - \frac1n\big)^{n} \to \frac1e < 1 \). The series converges.
Problem 5: \( \displaystyle\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2} \)
Integral test with \(t = \ln x\): \( \int_2^\infty \frac{dx}{x(\ln x)^2} = \int_{\ln 2}^\infty \frac{dt}{t^2} = \frac{1}{\ln 2} < \infty \). The series converges. Compare this with Example 12: a single extra power of the logarithm changes the answer.
Problem 6: \( \displaystyle\sum_{n=1}^{\infty} \frac{n!}{(a+1)(a+2)\cdots(a+n)} \), where \(a > 0\)
Ratio test: \( \frac{a_{n+1}}{a_n} = \frac{n+1}{a+n+1} \to 1 \), so no decision. Raabe:
\[ n\left(\frac{a_n}{a_{n+1}} - 1\right) = n\cdot\frac{a}{n+1} \longrightarrow a. \]
The series converges for \(a > 1\) and diverges for \(a < 1\). For \(a = 1\) the term is \( \frac{1}{n+1} \), the harmonic series (shifted), which diverges.
Final thoughts
Every test in this post is a variation on one idea: compare your series with one you already understand. Comparison tests do this directly; the ratio and root tests compare with a geometric series; Raabe and Gauss compare with p-series; and the integral test compares with an area. Once you see that common thread, the list of tests stops being a list to memorize and becomes a toolbox you know how to use.
Next in the series: series with terms of arbitrary sign, alternating series and Leibniz's test, and the difference between absolute and conditional convergence.