Sequences, Limits and the Idea of Convergence

What does it mean for a list of numbers to "get closer and closer" to something? The question sounds simple, but answering it precisely took mathematicians more than two thousand years, from Zeno's paradoxes in ancient Greece to the rigorous definitions of the nineteenth century. Limits are the foundation of calculus, and once you understand convergence properly, derivatives, integrals and infinite series all make far more sense.

What Is a Sequence?

A sequence is an ordered list of numbers \( a_1, a_2, a_3, \ldots \), usually given by a formula for the \( n \)th term or by a rule connecting each term to the previous one. Some examples:

• \( a_n = \dfrac{1}{n} \): \( 1, \tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \ldots \)
• \( a_n = (-1)^n \): \( -1, 1, -1, 1, \ldots \)
• \( a_n = n^2 \): \( 1, 4, 9, 16, \ldots \)

The first sequence settles down towards 0. The second keeps jumping between two values. The third grows without bound. These are the three basic behaviours: a sequence may converge to a limit, oscillate, or diverge to infinity.

The Precise Definition of a Limit

Saying "the terms get close to \( L \)" is too vague. The terms of \( 1/n \) get closer to \( -1 \) as well, in the sense that each is nearer than the last, yet \( -1 \) is clearly not the limit. The modern definition, developed by Augustin-Louis Cauchy in the 1820s and made fully rigorous by Karl Weierstrass later in the century, says:

The sequence \( (a_n) \) converges to \( L \) if for every \( \varepsilon > 0 \) there exists a natural number \( N \) such that

$$ |a_n - L| < \varepsilon \quad \text{for all } n > N $$

Think of it as a challenge game. Your opponent picks any tolerance \( \varepsilon \), however tiny. You must respond with a point \( N \) in the sequence beyond which every term lies within \( \varepsilon \) of \( L \). If you can always win, the limit is \( L \).

Worked Example 1: Proving a Limit

Prove that

$$ \lim_{n \to \infty} \frac{2n + 1}{n + 3} = 2 $$

Step 1: Compute the distance from the limit.

$$ \left| \frac{2n + 1}{n + 3} - 2 \right| = \left| \frac{2n + 1 - 2(n + 3)}{n + 3} \right| = \left| \frac{-5}{n + 3} \right| = \frac{5}{n + 3} $$

Step 2: Make it smaller than \( \varepsilon \). We need

$$ \frac{5}{n + 3} < \varepsilon \quad \Longleftrightarrow \quad n + 3 > \frac{5}{\varepsilon} \quad \Longleftrightarrow \quad n > \frac{5}{\varepsilon} - 3 $$

Step 3: Choose \( N \). Take \( N \) to be any natural number at least \( \dfrac{5}{\varepsilon} - 3 \). Then every \( n > N \) satisfies the inequality, completing the proof.

A concrete check: if \( \varepsilon = 0.01 \), we need \( n > 500 - 3 = 497 \). For \( n = 498 \), the term is \( 997/501 \approx 1.99002 \), which is indeed within 0.01 of 2.

Useful Limit Laws

Proving every limit from the definition would be exhausting. Fortunately, if \( a_n \to A \) and \( b_n \to B \), then

• \( a_n + b_n \to A + B \) and \( a_n b_n \to AB \),
• \( a_n / b_n \to A / B \) provided \( B \ne 0 \).

A handy technique for rational expressions is dividing top and bottom by the highest power of \( n \):

$$ \frac{3n^2 - n}{5n^2 + 4} = \frac{3 - 1/n}{5 + 4/n^2} \to \frac{3 - 0}{5 + 0} = \frac{3}{5} $$

The squeeze theorem is also powerful: if \( b_n \le a_n \le c_n \) and both outer sequences tend to \( L \), so does \( a_n \). For instance, \( -1/n \le \sin(n)/n \le 1/n \), so \( \sin(n)/n \to 0 \).

Worked Example 2: A Recursive Sequence

Define \( a_1 = \sqrt{2} \) and \( a_{n+1} = \sqrt{2 + a_n} \). The terms are

$$ \sqrt{2}, \quad \sqrt{2 + \sqrt{2}}, \quad \sqrt{2 + \sqrt{2 + \sqrt{2}}}, \ldots $$

which are approximately 1.414, 1.848, 1.962, 1.990, ...

Step 1: Convergence. One can show by induction that the sequence is increasing and always less than 2. The Monotone Convergence Theorem says every increasing sequence that is bounded above converges. So a limit \( L \) exists.

Step 2: Find the limit. Take limits on both sides of the rule. As \( n \to \infty \), both \( a_{n+1} \) and \( a_n \) tend to \( L \):

$$ L = \sqrt{2 + L} $$

Square both sides:

$$ L^2 = 2 + L \quad \Rightarrow \quad L^2 - L - 2 = 0 \quad \Rightarrow \quad (L - 2)(L + 1) = 0 $$

So \( L = 2 \) or \( L = -1 \). All terms are positive, so \( L = 2 \), matching the numerical values.

Series: Adding Infinitely Many Terms

An infinite series \( \sum a_n \) converges if its sequence of partial sums converges. The most important example is the geometric series:

$$ a + ar + ar^2 + \cdots = \frac{a}{1 - r}, \qquad |r| < 1 $$

Worked Example 3: Is 0.999... Equal to 1?

Write the recurring decimal as a series:

$$ 0.999\ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots $$

This is geometric with \( a = \tfrac{9}{10} \) and \( r = \tfrac{1}{10} \), so

$$ 0.999\ldots = \frac{9/10}{1 - 1/10} = \frac{9/10}{9/10} = 1 $$

The two numbers are exactly equal: the limit of the partial sums 0.9, 0.99, 0.999, ... is 1.

Not every series with shrinking terms converges. The harmonic series \( 1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4} + \cdots \) diverges. Around 1350 the French scholar Nicole Oresme proved this by grouping terms: \( \tfrac{1}{3} + \tfrac{1}{4} > \tfrac{1}{2} \), \( \tfrac{1}{5} + \cdots + \tfrac{1}{8} > \tfrac{1}{2} \), and so on, adding infinitely many halves.

A Famous Limit: The Number e

Consider \( a_n = \left(1 + \tfrac{1}{n}\right)^n \), which arises from compound interest:

• \( n = 1 \): 2
• \( n = 10 \): 2.5937...
• \( n = 100 \): 2.7048...
• \( n = 1000 \): 2.7169...

The terms increase towards \( e \approx 2.71828 \). Jacob Bernoulli studied this limit around 1683 while investigating continuously compounded interest.

Common Mistakes to Avoid

Students often lose marks on limits for the same few reasons:

• Treating infinity as a number. Expressions such as \( \infty - \infty \) or \( \infty / \infty \) are not automatically 0 or 1; they are indeterminate and need algebraic rearrangement first.
• Assuming shrinking terms mean a convergent series. The terms of a series must tend to 0 for it to converge, but that alone is not enough, as the harmonic series shows.
• Forgetting to prove a limit exists. In recursive sequences, solving \( L = f(L) \) only tells you what the limit must be if it exists. The sequence \( a_{n+1} = 2a_n \) with \( a_1 = 1 \) gives \( L = 2L \), so \( L = 0 \), yet the terms 1, 2, 4, 8, ... clearly diverge.
• Confusing the sequence with the series. The sequence \( \tfrac{1}{2^n} \) tends to 0, while the series \( \sum \tfrac{1}{2^n} \) sums to 1.

Did You Know?

• Zeno of Elea (around 450 BC) argued that a runner can never finish a race because they must first cover half the distance, then half of the remainder, forever. Convergent series resolve the paradox: \( \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots = 1 \).
• The harmonic series diverges incredibly slowly: you need more than 12,000 terms for the sum to pass 10.
• Archimedes summed a geometric series around 250 BC to find the area of a parabolic segment.

Key Takeaways

• A sequence can converge, oscillate or diverge.
• The epsilon-N definition makes "getting arbitrarily close" precise and provable.
• Limit laws, dividing by the highest power and the squeeze theorem make limits easier to compute.
• Bounded monotone sequences always converge; for recursive sequences, solve \( L = f(L) \).
• A series converges when its partial sums do; geometric series with \( |r| < 1 \) sum to \( a/(1 - r) \).

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