x̄ - > Proofs of sequences and series

Sequences and Series in Real Analysis

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1. Sequences

(a) Definition of Convergence

A sequence \( \{a_n\} \) converges to \( L \) if for every \( \varepsilon > 0 \), there exists \( N \in \mathbb{N} \) such that for all \( n > N \), \( |a_n - L| < \varepsilon \).

Example: The sequence \( a_n = \frac{1}{n} \) converges to 0 because for any \( \varepsilon > 0 \), choose \( N > \frac{1}{\varepsilon} \). Then, for all \( n > N \), \( |a_n - 0| = \frac{1}{n} < \varepsilon \).

Proof:

Let \( \varepsilon > 0 \). Choose \( N > \frac{1}{\varepsilon} \). Then, for \( n > N \),

\[ |a_n - 0| = \frac{1}{n} < \frac{1}{N} < \varepsilon. \]

Thus, \( a_n \to 0 \).

(b) Monotone Convergence Theorem

If \( \{a_n\} \) is a monotone (increasing or decreasing) sequence and is bounded, then it converges.

Example: The sequence \( a_n = 1 - \frac{1}{n} \) is increasing and bounded above by 1. Therefore, it converges to 1.

Proof:

Assume \( \{a_n\} \) is increasing and bounded above by \( M \). Define \( L = \sup \{a_n\} \). For any \( \varepsilon > 0 \), \( L - \varepsilon \) is not an upper bound, so there exists \( N \) such that \( a_N > L - \varepsilon \). For \( n > N \), \( a_n \leq L \). Hence,

\[ L - \varepsilon < a_n \leq L \quad \text{for all } n > N. \]

Thus, \( a_n \to L \).

(c) Bolzano-Weierstrass Theorem

Every bounded sequence has a convergent subsequence.

Example: The sequence \( a_n = (-1)^n \) is bounded but oscillatory. Its subsequences \( a_{2n} = 1 \) and \( a_{2n+1} = -1 \) converge to 1 and -1, respectively.

Proof:

Let \( \{a_n\} \) be bounded. Using the completeness property of \( \mathbb{R} \), consider the intervals that "trap" \( \{a_n\} \) progressively. A nested subsequence can be constructed, converging to a point in \( \mathbb{R} \).

(d) Cauchy Sequences

A sequence \( \{a_n\} \) is Cauchy if for every \( \varepsilon > 0 \), there exists \( N \) such that \( |a_m - a_n| < \varepsilon \) for all \( m, n > N \). In \( \mathbb{R} \), every Cauchy sequence converges.

Example: \( a_n = \frac{1}{n} \) is a Cauchy sequence because \( |a_m - a_n| < \frac{1}{N} \) for large \( N \).

Proof:

Let \( \{a_n\} \) be Cauchy. For any \( \varepsilon > 0 \), there exists \( N \) such that \( |a_m - a_n| < \varepsilon \) for all \( m, n > N \). This implies \( \{a_n\} \) is bounded. By Bolzano-Weierstrass, it has a convergent subsequence. Since \( \{a_n\} \) is Cauchy, the entire sequence converges to the same limit.

(e) Limits Superior and Inferior

Let \( \{a_n\} \) be a sequence. The limit superior (\( \limsup a_n \)) is the greatest limit of all subsequences, and the limit inferior (\( \liminf a_n \)) is the smallest limit.

Example: For \( a_n = (-1)^n + \frac{1}{n} \), \( \limsup a_n = 1 \) and \( \liminf a_n = -1 \).

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2. Series

(a) Infinite Series and Convergence

An infinite series \( \sum_{n=1}^\infty a_n \) converges if the sequence of partial sums \( S_N = \sum_{n=1}^N a_n \) converges.

Proof:

Let \( \{S_N\} \) converge to \( L \). For any \( \varepsilon > 0 \), there exists \( N \) such that for \( m, n > N \),

\[ |S_m - S_n| = \left| \sum_{k=n+1}^m a_k \right| < \varepsilon. \]

Thus, the series converges.

(b) Convergence Tests

• Comparison Test: If \( 0 \leq a_n \leq b_n \) and \( \sum b_n \) converges, then \( \sum a_n \) converges.
• Ratio Test: If \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), then \( \sum a_n \) converges.
• Root Test: If \( \lim_{n \to \infty} \sqrt[n]{|a_n|} < 1 \), then \( \sum a_n \) converges.

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