Analysis

Numerical series and convergence tests

Infinite sums of numbers and the criteria that decide whether they add up to a finite value.

IntuitionIntuition: adding infinitely many terms

Walk halfway to a wall, then halfway again, then halfway again, forever: the distances you cover are $1/2$ , $1/4$ , $1/8,\dots$ , and even though you take infinitely many steps, the total distance never exceeds the width of the room — it approaches exactly $1$ (in units of the room's width). A numerical series is precisely this idea made rigorous: an infinite sum $\sum_{n=1}^{\infty} a_n$ of numbers, and the central question is whether such a sum settles down to a finite value or grows without bound.

Picture each term of the harmonic series

1,\ 1/2,\ 1/3,\dots,\ 1/n

as the height of a bar of width

1

: the bars shown here for

n

up to 12 accumulate to the partial sum

S_n

, and comparing their total area to the area under the curve

y=1/x

is exactly the geometric idea behind the integral test.

UndergraduateDefinition: series and partial sums

Definition: Numerical series and convergence

Given a sequence of real numbers $a_n$ , the $n$ -th partial sum is $S_N = \sum_{n=1}^{N} a_n$ , the sum of the first $n$ terms. The infinite series $\sum_{n=1}^{\infty} a_n$ is said to converge if the sequence of partial sums $S_N$ has a finite limit as $N\to\infty$ ; otherwise the series diverges.

S_N = \sum_{n=1}^{N} a_n

Here $a_n$ denotes the general term of the series (a function of the index $n$ ), and $S_N$ is a genuine, finite sum for every fixed $n$ — it only becomes infinite in the limit. This is the key conceptual jump: an infinite series is not itself a sum in the ordinary sense but the limit of a sequence of ordinary, finite sums.

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

A necessary (but not sufficient) condition for convergence follows immediately: if $\sum_{n=1}^{\infty} a_n$ converges then $a_n \to 0$ , since consecutive partial sums $S_N$ must get arbitrarily close together. The word 'necessary' matters — as the pitfall below shows, $a_n \to 0$ alone never guarantees convergence.

UndergraduateStandard convergence tests

Checking the definition of convergence directly, via the limit of $S_N$ , is often impossible in closed form. Instead, mathematicians have developed a toolbox of tests that decide convergence from the shape of $a_n$ alone, summarized below.

Comparing the standard convergence tests
Test	Condition	Example
Geometric series	Converges iff $\|r\|<1$	$\sum_{n=0}^{\infty} ar^n$
p-series	Converges iff $p>1$	$\sum_{n=1}^{\infty} \dfrac{1}{n^p}$
Ratio test	Converges if $L<1$ , diverges if $L>1$	$L=\lim_{n\to\infty}\left\|\dfrac{a_{n+1}}{a_n}\right\|$
Root test	Converges if $L<1$ , diverges if $L>1$	$L=\limsup_{n\to\infty}\sqrt[n]{\|a_n\|}$
Leibniz (alternating)	Converges if $a_n \ge a_{n+1} \ge 0,\ a_n\to0$	$\sum_{n=1}^{\infty}(-1)^{n-1}a_n$

The ratio test compares consecutive terms directly: compute $L=\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$ ; if $L<1$ the series converges absolutely, if $L>1$ it diverges, and if $L$ equals $1$ the test is inconclusive. It is most effective when $a_n$ involves factorials or $n$ -th powers, since those simplify nicely in a ratio.

The root test uses $L=\limsup_{n\to\infty}\sqrt[n]{|a_n|}$ instead, with the same decision rule ( $L<1$ converges, $L>1$ diverges); it is most effective when $a_n$ itself is an $n$ -th power. The Leibniz test handles alternating series $\sum_{n=1}^{\infty}(-1)^{n-1}a_n$ directly: if the positive terms $a_n$ decrease monotonically to $0$ , the series converges even when it is not absolutely convergent, as for the alternating harmonic series.

UndergraduateKey theorems: the geometric series and the integral test

Theorem: Geometric series theorem

For real numbers $a$ and $r$ with $a$ nonzero, the geometric series $\sum_{n=0}^{\infty} ar^n$ converges if and only if $|r|<1$ , and in that case $\sum_{n=0}^{\infty} ar^n = \dfrac{a}{1-r}$ .

Why is it true?

The geometric series is the one series whose partial sums can be written in closed form, so it is both the simplest possible convergence criterion and the yardstick against which many other tests (ratio, root, comparison) are calibrated: they all work by comparing a general series to a geometric one.

Proof

Step 1 (write the partial sum in closed form). Multiply $S_N$ by $r$ : $S_N=\sum_{n=0}^{N-1}ar^n$ and $rS_N=\sum_{n=0}^{N-1}ar^{n+1}=\sum_{n=1}^{N}ar^n$ . Subtracting, almost every term cancels: $S_N-rS_N=a-ar^N$ , so $(1-r)S_N=a(1-r^N)$ .

Step 2 (solve for the partial sum). If $r\neq1$ this gives $S_N=a\dfrac{1-r^N}{1-r}$ , an exact, finite formula for every $n$ — no limit has been taken yet.

Step 3 (take the limit). If $|r|<1$ , then $r^N\to0$ as $N\to\infty$ (a power with base of absolute value less than $1$ shrinks to zero), so $S_N\to a\dfrac{1-0}{1-r}=\dfrac{a}{1-r}$ , which is exactly $\sum_{n=0}^{\infty} ar^n = \dfrac{a}{1-r}$ .

Step 4 (the converse: divergence). If $|r|\ge1$ and $r\neq1$ , then $r^N$ does not tend to any finite limit (it oscillates or grows without bound), so $S_N$ has no limit and the series diverges; if $r=1$ then $S_N$ equals $Na$ , which diverges to infinity since $a$ is nonzero. This covers every case, proving the 'if and only if'.

Theorem: Integral test

Let $f(x)$ be a positive, decreasing, continuous function on $[1,\infty)$ with $a_n$ equal to $f(n)$ for every integer $n$ . Then the series $\sum_{n=1}^{\infty} a_n$ converges if and only if the improper integral $\int_1^{\infty} f(x)\,dx$ converges.

Why is it true?

Many series, such as the $p$ -series, have no closed-form partial sum, so a direct limit computation is out of reach; but if the terms come from a smooth decreasing function, the discrete sum and the continuous integral track each other closely, and integrals are usually far easier to evaluate or bound.

Proof

Step 1 (bound each term by a strip of the integral). Since $f(x)$ is decreasing, for every integer $n\ge1$ and $x\in[n,n+1]$ we have $f(n+1)\le f(x)\le f(n)$ . Integrating over that unit interval gives $f(n+1)\le\int_n^{n+1}f(x)\,dx\le f(n)$ .

Step 2 (sum the bounds). Summing the right-hand inequality for $n=1,\dots,N$ gives $\int_1^{N+1}f(x)\,dx\le\sum_{n=1}^{N}a_n=S_N$ ; summing the left-hand inequality for $n=1,\dots,N-1$ gives $S_N-a_1\le\int_1^{N}f(x)\,dx$ . So the partial sum $S_N$ is sandwiched between two integrals over intervals that both grow like $[1,N]$ .

Step 3 (convergence transfers one way). If $\int_1^{\infty} f(x)\,dx$ converges, then $\int_1^{N}f(x)\,dx$ is bounded above as $N\to\infty$ , so by Step 2 the increasing sequence $S_N$ is bounded above, and a bounded increasing sequence of real numbers always converges (a form of the monotone convergence property of $\mathbb{R}$ ); hence $\sum_{n=1}^{\infty} a_n$ converges.

Step 4 (divergence transfers the other way). Conversely, if $\int_1^{\infty} f(x)\,dx$ diverges (grows without bound), then $\int_1^{N+1}f(x)\,dx\to\infty$ as $N\to\infty$ , and by the first inequality in Step 2, $S_N$ is bounded below by a quantity tending to infinity, so $S_N$ also diverges to infinity. Combining Steps 3 and 4 gives the full 'if and only if'.

UndergraduateReal-World Applications and Worked Examples

Series with a closed-form sum are the backbone of financial mathematics: the present value of a perpetuity or a loan's amortization schedule is a geometric series. In computer science, the running time of divide-and-conquer algorithms and the analysis of recursive data structures often reduce to bounding a series. In physics, damped oscillations and the decay of successive echoes in a resonant cavity form geometric series, and Taylor and Fourier coefficients in signal processing are judged by exactly the tests introduced above (ratio test for radius of convergence, comparison to a p-series for decay rate). In biology, discrete population models with constant per-generation growth or decline are literally geometric series.

Example: Present value of a perpetuity

An investment pays $100$ at the end of every year, forever, and the annual discount rate is $10\%$ , so a payment received $n$ years from now is worth $100(1.1)^{-n}$ today (all amounts in the same currency unit). Compute the total present value of all future payments.

Solution

Step 1: write the total present value as a series. Summing the discounted payments for $n=1,2,3,\dots$ gives $V=\sum_{n=1}^{\infty}100(1.1)^{-n}$ .

Step 2: recognize the geometric series. This is $\sum_{n=0}^{\infty} ar^n$ with $a=100/1.1$ and $r=1/1.1$ (reindexing to start at $n=0$ ), and $|r|=1/1.1<1$ , so the theorem applies.

Step 3: apply the closed-form sum. Using $\sum_{n=0}^{\infty} ar^n = \dfrac{a}{1-r}$ gives $V=\dfrac{100/1.1}{1-1/1.1}=\dfrac{100/1.1}{0.1/1.1}=\dfrac{100}{0.1}=1000$ .

Step 4: interpret. The perpetuity is worth exactly $1000$ today, even though it pays out forever — a direct real-world instance of an infinite series summing to a finite value.

Example: Ratio test on a factorial series

Determine whether the series $\sum_{n=1}^{\infty}\dfrac{2^n}{n!}$ converges, using the ratio test.

Solution

Step 1: write $a_n=\dfrac{2^n}{n!}$ and $a_{n+1}=\dfrac{2^{n+1}}{(n+1)!}$ .

Step 2: form the ratio. $\dfrac{a_{n+1}}{a_n}=\dfrac{2^{n+1}}{(n+1)!}\cdot\dfrac{n!}{2^n}=\dfrac{2}{n+1}$ , using $(n+1)!=(n+1)\cdot n!$ .

Step 3: take the limit $L=\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$ : here $L=\lim_{n\to\infty}\dfrac{2}{n+1}=0$ .

Step 4: conclude. Since $L=0<1$ , the ratio test guarantees the series converges (in fact absolutely) — the factorial in the denominator eventually overwhelms any fixed exponential in the numerator.

For $|r|<1$ , what does the geometric series $\sum_{n=0}^{\infty} ar^n$ converge to?

For the $p$ -series $\sum_{n=1}^{\infty} \dfrac{1}{n^p}$ , for which values of $p$ does it converge?

A series has terms satisfying $a_n \to 0$ . What can be concluded about whether the series converges?

A bank account pays a fixed nominal interest, and an analyst models the present value of an indefinitely long stream of equal future cash flows. Which mathematical object are they implicitly summing?