Infinite Series

12 min read

Textbook Reference

Series build directly on sequences, especially the meaning of a limit and the Monotonic Sequence Theorem.

Key idea

How do you add infinitely many numbers? You do not. You add finitely many, watch where the running total heads, and call that destination the sum.

Zeno’s old puzzle is the picture. To cross a room you walk half the distance, then half of what remains, then half again, forever. The distances are $\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{8}, \dots$. Zeno claimed you could never arrive because there are infinitely many steps. But you do arrive: the running totals $\tfrac{1}{2}, \tfrac{3}{4}, \tfrac{7}{8}, \tfrac{15}{16}, \dots$ march steadily toward $1$, the width of the room. The “infinite sum” is just the limit those running totals approach.

That is the entire idea, and it converts a brand-new question into one you already solved. Adding up a series means building the sequence of running totals (called partial sums) and asking the question from the sequence lessons: does this sequence have a limit? If it does, that limit is the sum, and the series converges. If it does not, the series diverges. A series is not a new kind of object; it is a sequence of partial sums wearing a plus sign.

Prerequisite Check

• Know what a sequence is and read $\{a_n\}$ notation (first sequence lesson)
• Find the limit of a sequence, including by dividing by the highest power (second sequence lesson)
• State the Monotonic Sequence Theorem (third sequence lesson)
• Decompose a fraction by partial fractions, for example $\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$
• Read sigma notation $\displaystyle\sum_{n=1}^{\infty} a_n$ as a sum of terms

If these are solid, you are ready.

Quick Reference

The series. Given a sequence $\{a_n\}$, the infinite series is the formal sum \[ \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots \]

Partial sums. The $n$th partial sum is the running total of the first $n$ terms, \[ s_n = \sum_{i=1}^{n} a_i = a_1 + a_2 + \cdots + a_n. \]

Definition of the sum (formal gloss). If the sequence of partial sums $\{s_n\}$ has a finite limit $S$, the series converges and its sum is \[ \sum_{n=1}^{\infty} a_n = \lim_{n\to\infty} s_n = S. \] (In plain words: the sum is the number the running totals close in on.) If $\{s_n\}$ has no finite limit, the series diverges.

The two sequences. Every series carries two sequences: the terms $\{a_n\}$ and the partial sums $\{s_n\}$. They are different. The sum is the limit of the partial sums, not of the terms.

Procedure (when a partial-sum formula is findable).

1. Write the $n$th partial sum $s_n$ in closed form (telescoping and geometric series allow this).
2. Take $\displaystyle\lim_{n\to\infty} s_n$.
3. If the limit is a finite number, that is the sum; otherwise the series diverges.

Caution. A finite number of terms at the front never changes whether a series converges or diverges; it only shifts the sum.

Key Concepts

1. A Series Is the Limit of Its Partial Sums

The leap from a sequence to a series is the leap from a list of numbers to their running total. Take the Zeno distances $a_n = \dfrac{1}{2^n}$. The partial sums are \[ s_1 = \tfrac{1}{2}, \quad s_2 = \tfrac{3}{4}, \quad s_3 = \tfrac{7}{8}, \quad s_4 = \tfrac{15}{16}, \dots \] In closed form $s_n = 1 - \dfrac{1}{2^n}$, and so $\displaystyle\lim_{n\to\infty} s_n = 1$. That limit is the sum: \[ \sum_{n=1}^{\infty} \frac{1}{2^n} = 1. \]

The same machine handles any series. Build $\{s_n\}$, take the limit. The hard part in practice is finding a closed form for $s_n$; only special families (telescoping and geometric) hand one over directly.

2. Reading the Sum Off a Partial-Sum Formula

When the partial sum is handed to you, finding the sum is a sequence-limit problem and nothing more.

Example 1 (partial sum given). Suppose the partial sums of $\displaystyle\sum_{n=1}^{\infty} a_n$ are $s_n = \dfrac{2n}{3n+5}$. Find the sum of the series.

Goal. The sum is $\lim_{n\to\infty} s_n$. Compute it like any sequence limit.

Work. \[ \sum_{n=1}^{\infty} a_n = \lim_{n\to\infty} \frac{2n}{3n+5} = \lim_{n\to\infty} \frac{2}{3 + \frac{5}{n}} = \frac{2}{3}. \]

Answer: $\dfrac{2}{3}$

Recap. A series is its sequence of partial sums in disguise. Once $s_n$ is known, the sum is a limit you already know how to take.

3. Telescoping Series

A telescoping series is one whose partial sum collapses because consecutive terms cancel. The trick is to split each term into a difference, often by partial fractions, so the middle of the sum folds up.

Example 2 (telescoping sum). Show that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ converges, and find its sum.

Goal. Split each term into a difference, write out the partial sum, watch the middle cancel, then take the limit.

Work. Partial fractions give \[ \frac{1}{i(i+1)} = \frac{1}{i} - \frac{1}{i+1}. \] So the partial sum is \[ s_n = \sum_{i=1}^{n} \left( \frac{1}{i} - \frac{1}{i+1} \right) = \left(1 - \tfrac{1}{2}\right) + \left(\tfrac{1}{2} - \tfrac{1}{3}\right) + \cdots + \left(\tfrac{1}{n} - \tfrac{1}{n+1}\right). \] Every interior term cancels with its neighbor, leaving only the first and last pieces: \[ s_n = 1 - \frac{1}{n+1}. \] Taking the limit, \[ \lim_{n\to\infty} s_n = 1 - 0 = 1. \]

Answer: the series converges and $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1$.

Recap. The cancellation collapses the sum the way a folding telescope collapses, which is where the name comes from. Notice the two sequences in play: the terms $a_n = \dfrac{1}{n(n+1)} \to 0$, while the partial sums $s_n \to 1$. Those are different limits, and the sum is the second one.

(Common error: writing the cancellation without keeping the surviving end terms. After everything in the middle cancels, the first term $1$ and the final $-\dfrac{1}{n+1}$ remain. Track which pieces survive before taking the limit.)

4. The Harmonic Series: Terms Going to Zero Is Not Enough

The harmonic series is the chapter’s central cautionary case: its terms shrink to zero, yet it diverges.

Example 3 (the harmonic series diverges). Show that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$ diverges.

Goal. Group the terms in blocks whose lengths double, and show each block adds more than $\tfrac{1}{2}$, so the partial sums grow without bound.

Work. Consider the partial sums at positions $2, 4, 8, 16, \dots$ and bound each block from below by replacing every term in it with the smallest term in that block: \[ s_4 = 1 + \tfrac{1}{2} + \left(\tfrac{1}{3} + \tfrac{1}{4}\right) > 1 + \tfrac{1}{2} + \left(\tfrac{1}{4} + \tfrac{1}{4}\right) = 1 + \tfrac{2}{2}, \] \[ s_8 > 1 + \tfrac{1}{2} + \left(\tfrac{1}{4} + \tfrac{1}{4}\right) + \left(\tfrac{1}{8} + \tfrac{1}{8} + \tfrac{1}{8} + \tfrac{1}{8}\right) = 1 + \tfrac{3}{2}. \] Continuing the pattern, $s_{2^k} > 1 + \dfrac{k}{2}$. Since $1 + \dfrac{k}{2} \to \infty$, the partial sums $\{s_n\}$ are unbounded.

Answer: the partial sums grow without bound, so the harmonic series diverges.

Recap. The terms $\dfrac{1}{n}$ shrink to $0$, yet the sum is infinite. Shrinking terms are necessary but not sufficient for a series to converge: they keep the running total from jumping, but if they shrink too slowly the total can still creep off to infinity.

(Common error: assuming that because $a_n \to 0$ the series must converge. The harmonic series is the standard counterexample. “Terms go to zero” rules nothing in; it only fails to rule a series out.)

Inline Self-Check

Question. A series has partial sums $s_n = 3 - \dfrac{1}{2^n}$. Does the series converge, and if so, what is its sum?

Common Errors Summary

Common Misconceptions

Leveled Practice

Attempt each problem before opening the answer.

Level 1: Direct Application

Problem 1. The partial sums of a series are $s_n = 2 - 3(0.8)^n$. Find the sum of the series.

Problem 2. The partial sums of a series are $s_n = \dfrac{n^2 - 1}{4n^2 + 1}$. Find the sum.

Problem 3. If the $n$th partial sum of $\displaystyle\sum_{i=1}^{\infty} a_i$ is $s_n = \dfrac{n-1}{n+1}$, find $a_4$ and the sum of the series.

Level 2: Telescoping

Problem 4. Determine whether $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}\right)$ converges, and if so find its sum.

Problem 5. Find the sum of $\displaystyle\sum_{n=1}^{\infty}\frac{3}{n(n+3)}$.

Level 3: Reasoning

Problem 6. A series of positive terms has partial sums that satisfy $s_n \le 1000$ for every $n$. Explain why the series must converge.

Problem 7. Both the terms of a series go to $0$ and the partial sums grow without bound. Is this a contradiction? Name a series where it happens.

Mastery Checklist

You have mastered this skill when you can do all of the following without notes:

• Define the $n$th partial sum $s_n$ of a series
• State that the sum of a series is the limit of its partial sums
• Find the sum when a closed form for $s_n$ is available
• Recognize and evaluate a telescoping series
• Distinguish the sequence of terms from the sequence of partial sums
• State that the harmonic series diverges even though its terms go to $0$
• Use the Monotonic Sequence Theorem to argue a positive series with bounded partial sums converges

Mental Model

Think of pouring water into a tall glass, one splash at a time. Each splash is a term $a_n$. The water level after $n$ splashes is the partial sum $s_n$.

A series converges when the water level rises toward a fixed mark and stops there, even as you keep adding ever-smaller splashes. That mark is the sum. A series diverges when the level never settles: it either overflows (rises without bound, like the harmonic series) or sloshes up and down without homing in on one mark.

The key warning is in this picture too. Smaller and smaller splashes ($a_n \to 0$) keep the level from jumping, but they do not guarantee the glass stops filling. Pour slowly enough forever and the glass can still overflow. That is the harmonic series, and it is why a separate convergence test is needed for almost every series you meet.

Connections

Built From

• Sequences and notation: the terms $\{a_n\}$ are a sequence, and the partial sums $\{s_n\}$ are a second sequence built from them.
• Limits of sequences: the sum of a series is defined as a limit of the partial-sum sequence, so every tool for evaluating a sequence limit applies directly.
• Properties of sequences: for a positive-term series, the partial sums are increasing, so the Monotonic Sequence Theorem decides convergence by boundedness.

Leads To

• The divergence test: the immediate consequence that if $\sum a_n$ converges then $a_n \to 0$, used in its contrapositive form to rule series out.
• Geometric and telescoping series: the two families whose partial sums have closed forms, so their sums can be computed exactly.
• The integral test and the comparison tests: for series whose partial sums have no closed form, these tests decide convergence by other means, all resting on this partial-sum definition.

Why It Matters

A convergent series is how an exact quantity is built from infinitely many pieces: the decimal expansion of a number, the value of a function from its power series, the steady-state level of a repeated dose of medication, the total distance traveled by a bouncing ball. In computing, a series is summed term by term until the partial sum stops changing to machine precision, which is the partial-sum definition turned into a stopping rule. Knowing that the partial sums actually converge is what makes that computation meaningful.

Audience Notes

For students who find math intimidating: you are not adding infinitely many numbers at once. You add a few, see where the running total is heading, and that destination is the answer. It is the same “where is this going” question you asked about sequences.

For students who want depth: the harmonic series is worth real attention. It shows that the necessary condition ($a_n \to 0$) and the actual conclusion (convergence) are genuinely different, which is the gap that the rest of the chapter exists to fill.

For students aimed at a career in computing or engineering: summing a series numerically means accumulating partial sums until they stabilize. The harmonic series is a cautionary tale here too: a naive “stop when the next term is tiny” rule would wrongly report it as convergent.

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