Real Analysis: Sequences, Series and Continuity

Real analysis is the course where calculus stops relying on intuition and starts relying on proof. Ideas you have used for years, such as limits, infinite sums and continuous functions, are given precise definitions, and every result is justified step by step. It can feel abstract at first, but once the definitions click, the subject becomes logical and even elegant.

Why Rigour Matters

Intuition can mislead. The series \( 1 - 1 + 1 - 1 + \dots \) seems to equal 0 if you group terms one way and 1 if you group them another; in fact it does not converge at all. The harmonic series \( 1 + \frac{1}{2} + \frac{1}{3} + \dots \) has terms shrinking to zero yet grows without bound. Real analysis gives us definitions precise enough to settle such questions for certain.

The foundation is the completeness axiom of the real numbers: every non-empty set of reals that is bounded above has a least upper bound, called its supremum. The rational numbers fail this property. For example, the set of rationals whose square is less than 2 has no rational least upper bound, because \( \sqrt{2} \) is irrational. Completeness is what makes limits behave well in \( \mathbb{R} \).

Sequences and Their Limits

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

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

In words: however small a tolerance \( \varepsilon \) you choose, the sequence eventually stays within that tolerance of \( L \) forever. The proof always has the same shape: start with an arbitrary \( \varepsilon \), work out how large \( n \) must be, and then choose \( N \) accordingly.

Worked Example 1: An Epsilon-N Proof

Prove that \( \displaystyle \lim_{n \to \infty} \frac{3n + 1}{n + 2} = 3 \).

Step 1 (rough work). Simplify the distance from the limit:

$$ \left| \frac{3n + 1}{n + 2} - 3 \right| = \left| \frac{3n + 1 - 3n - 6}{n + 2} \right| = \frac{5}{n + 2} $$

Step 2. We want \( \frac{5}{n + 2} < \varepsilon \), which holds when \( n + 2 > \frac{5}{\varepsilon} \), that is \( n > \frac{5}{\varepsilon} - 2 \).

Step 3 (the proof). Let \( \varepsilon > 0 \). Choose a natural number \( N \ge \frac{5}{\varepsilon} - 2 \). Then for all \( n > N \),

$$ \left| \frac{3n + 1}{n + 2} - 3 \right| = \frac{5}{n + 2} < \frac{5}{N + 2} \le \varepsilon $$

so the sequence converges to 3.

Concrete check: for \( \varepsilon = 0.01 \) we need \( n > 498 \). At \( n = 499 \), the distance is \( \frac{5}{501} \approx 0.00998 \), just inside the tolerance.

Useful Theorems About Sequences

• Uniqueness: a convergent sequence has exactly one limit.
• Boundedness: every convergent sequence is bounded (the converse is false: \( (-1)^n \) is bounded but diverges).
• Algebra of limits: limits of sums, products and quotients behave as expected, provided denominators do not tend to zero.
• Squeeze theorem: if \( a_n \le b_n \le c_n \) and both outer sequences tend to \( L \), so does \( b_n \).
• Monotone convergence theorem: a bounded, monotone sequence converges.
• Bolzano-Weierstrass theorem: every bounded sequence has a convergent subsequence.

Worked Example 2: A Recursive Sequence

Let \( a_1 = 1 \) and \( a_{n+1} = \sqrt{2 + a_n} \). Show that \( (a_n) \) converges and find its limit.

Step 1 (compute terms). \( a_2 = \sqrt{3} \approx 1.732 \), \( a_3 = \sqrt{3.732} \approx 1.932 \), \( a_4 \approx 1.983 \). The sequence appears to increase towards 2.

Step 2 (bounded above by 2, by induction). \( a_1 = 1 < 2 \). If \( a_n < 2 \), then \( a_{n+1} = \sqrt{2 + a_n} < \sqrt{4} = 2 \).

Step 3 (increasing). For \( 0 < a_n < 2 \), we have \( a_{n+1}^2 - a_n^2 = 2 + a_n - a_n^2 = (2 - a_n)(1 + a_n) > 0 \), so \( a_{n+1} > a_n \).

Step 4. By the monotone convergence theorem, the sequence converges to some \( L \).

Step 5 (find \( L \)). Take limits on both sides of the recurrence: \( L = \sqrt{2 + L} \), so \( L^2 - L - 2 = 0 \), giving \( (L - 2)(L + 1) = 0 \). Since all terms are positive, \( L = 2 \).

Note the order: we proved convergence first. Taking limits in a recurrence without knowing a limit exists can give nonsense.

Series

An infinite series \( \sum a_n \) converges if its sequence of partial sums \( S_n = a_1 + \dots + a_n \) converges. Two essential examples:

• Geometric series: \( \sum r^n \) converges if and only if \( |r| < 1 \).
• p-series: \( \sum \frac{1}{n^p} \) converges if and only if \( p > 1 \). So \( \sum \frac{1}{n^2} \) converges, while the harmonic series \( \sum \frac{1}{n} \) diverges.

The divergence test says that if \( a_n \not\to 0 \), the series diverges. The converse fails, as the harmonic series shows.

Convergence Tests

• Comparison test: if \( 0 \le a_n \le b_n \) and \( \sum b_n \) converges, then \( \sum a_n \) converges.
• Ratio test: if \( \left|\frac{a_{n+1}}{a_n}\right| \to \rho \), the series converges absolutely when \( \rho < 1 \), diverges when \( \rho > 1 \), and the test is inconclusive when \( \rho = 1 \).
• Alternating series test: if \( b_n \) decreases to 0, then \( \sum (-1)^n b_n \) converges.

Worked Example 3: The Ratio Test

Determine whether \( \displaystyle \sum_{n=1}^{\infty} \frac{n^2}{2^n} \) converges.

Step 1. Form the ratio:

$$ \frac{a_{n+1}}{a_n} = \frac{(n + 1)^2}{2^{n+1}} \cdot \frac{2^n}{n^2} = \frac{1}{2}\left( \frac{n + 1}{n} \right)^2 $$

Step 2. Take the limit: \( \frac{n + 1}{n} \to 1 \), so the ratio tends to \( \rho = \frac{1}{2} \).

Step 3. Since \( \rho < 1 \), the series converges. (Its sum is in fact 6.)

The alternating harmonic series \( 1 - \frac{1}{2} + \frac{1}{3} - \dots \) converges to \( \ln 2 \), but only conditionally: the series of absolute values diverges. Riemann showed that such a series can be rearranged to sum to any real number at all.

Continuity

A function \( f \) is continuous at \( c \) if for every \( \varepsilon > 0 \) there exists \( \delta > 0 \) such that

$$ |x - c| < \delta \implies |f(x) - f(c)| < \varepsilon $$

Small changes in input produce small changes in output. Equivalently, \( f \) is continuous at \( c \) when \( f(x_n) \to f(c) \) for every sequence \( x_n \to c \), which links continuity back to sequences.

Worked Example 4: An Epsilon-Delta Proof

Prove that \( f(x) = x^2 \) is continuous at \( x = 2 \).

Step 1. Factorise: \( |x^2 - 4| = |x - 2|\,|x + 2| \).

Step 2. Control \( |x + 2| \). If we insist \( \delta \le 1 \), then \( |x - 2| < 1 \) means \( 1 < x < 3 \), so \( |x + 2| < 5 \).

Step 3. Then \( |x^2 - 4| < 5|x - 2| \), which is less than \( \varepsilon \) when \( |x - 2| < \frac{\varepsilon}{5} \).

Step 4. Choose \( \delta = \min\left(1, \frac{\varepsilon}{5}\right) \). Whenever \( |x - 2| < \delta \), both conditions hold and \( |x^2 - 4| < \varepsilon \).

Big Theorems on Closed Intervals

For a function continuous on \( [a, b] \):

• Intermediate value theorem: \( f \) takes every value between \( f(a) \) and \( f(b) \). For example, \( g(x) = x^3 - x - 1 \) has \( g(1) = -1 \) and \( g(2) = 5 \), so it has a root between 1 and 2.
• Extreme value theorem: \( f \) attains a maximum and a minimum.
• Uniform continuity: a single \( \delta \) works for every point of the interval.

Did You Know?

• Augustin-Louis Cauchy and Karl Weierstrass, working in the 1800s, turned vague ideas of "infinitely small" quantities into the modern epsilon definitions.
• Weierstrass constructed a function that is continuous everywhere but differentiable nowhere, shocking mathematicians of his time.
• The harmonic series diverges so slowly that its first million terms add up to only about 14.4.

Key Takeaways

• Real analysis rests on the completeness of the real numbers.
• Convergence of a sequence is defined with \( \varepsilon \) and \( N \); proofs work backwards from the inequality to choose \( N \).
• Bounded monotone sequences converge, which is powerful for recursive sequences.
• Series convergence uses partial sums and tests such as comparison, ratio and alternating series.
• Continuity is defined with \( \varepsilon \) and \( \delta \) and gives the intermediate and extreme value theorems on closed intervals.

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