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Real Analysis Sequences and Series cheat sheet - grade college

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Math Grade college

Real Analysis Sequences and Series Cheat Sheet

A printable reference covering sequence limits, convergence tests, Cauchy criteria, power series, and absolute convergence for college.

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Real analysis sequences and series study how ordered lists of numbers and infinite sums behave as their indices grow without bound. This cheat sheet helps students organize the main definitions, limit laws, and convergence tests used in proof-based calculus and analysis. It is especially useful for comparing similar tests and remembering the exact conditions needed before applying them.

The core ideas are convergence, divergence, boundedness, monotonicity, and Cauchy behavior. For series, the most important distinction is between convergence of the terms an→0a_n \to 0 and convergence of the infinite sum ∑n=1∞an\sum_{n=1}^{\infty} a_n. Power series add the ideas of radius and interval of convergence, usually found with the ratio or root test.

Many proofs depend on precise quantifiers, especially the ε\varepsilon definition of a limit and the Cauchy criterion.

Key Facts

  • A sequence (an)(a_n) converges to LL if for every ε>0\varepsilon > 0 there exists N∈NN \in \mathbb{N} such that n≥Nn \ge N implies ∣an−L∣<ε|a_n - L| < \varepsilon.
  • Every convergent sequence is bounded, but a bounded sequence does not have to converge.
  • A monotone increasing sequence that is bounded above converges, and a monotone decreasing sequence that is bounded below converges.
  • A series ∑n=1∞an\sum_{n=1}^{\infty} a_n converges if and only if its partial sums sn=∑k=1naks_n = \sum_{k=1}^{n} a_k converge.
  • If ∑n=1∞an\sum_{n=1}^{\infty} a_n converges, then an→0a_n \to 0, but an→0a_n \to 0 alone does not guarantee that ∑n=1∞an\sum_{n=1}^{\infty} a_n converges.
  • The geometric series satisfies ∑n=0∞arn=a1−r\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} when ∣r∣<1|r| < 1 and diverges when ∣r∣≥1|r| \ge 1.
  • The pp-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges if p>1p > 1 and diverges if p≤1p \le 1.
  • For a power series ∑n=0∞cn(x−a)n\sum_{n=0}^{\infty} c_n(x-a)^n, the radius of convergence is often found from R=1lim sup⁡n→∞∣cn∣nR = \frac{1}{\limsup_{n\to\infty} \sqrt[n]{|c_n|}}.

Vocabulary

Sequence
A sequence is a function from the natural numbers to the real numbers, usually written as (an)(a_n).
Limit
The limit LL of a sequence is the number that the terms ana_n approach as n→∞n \to \infty.
Partial Sum
The nnth partial sum of a series is sn=∑k=1naks_n = \sum_{k=1}^{n} a_k, the finite sum of the first nn terms.
Absolute Convergence
A series ∑an\sum a_n converges absolutely if the positive series ∑∣an∣\sum |a_n| converges.
Conditional Convergence
A series ∑an\sum a_n converges conditionally if ∑an\sum a_n converges but ∑∣an∣\sum |a_n| diverges.
Radius of Convergence
The radius of convergence RR is the distance from the center aa within which a power series ∑cn(x−a)n\sum c_n(x-a)^n converges.

Common Mistakes to Avoid

  • Using an→0a_n \to 0 as a convergence test for ∑an\sum a_n is wrong because it is only a necessary condition, not a sufficient condition.
  • Applying the ratio test when the limit equals 11 is wrong because lim⁡n→∞∣an+1an∣=1\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = 1 gives no conclusion.
  • Forgetting endpoint checks in power series is wrong because the ratio or root test usually gives convergence only for ∣x−a∣<R|x-a| < R and divergence for ∣x−a∣>R|x-a| > R.
  • Assuming bounded means convergent is wrong because sequences such as an=(−1)na_n = (-1)^n are bounded but do not approach one limit.
  • Confusing absolute and conditional convergence is wrong because ∑∣an∣\sum |a_n| converging implies ∑an\sum a_n converges, but the reverse is not always true.

Practice Questions

  1. 1 Determine whether the sequence an=3n2−12n2+5a_n = \frac{3n^2 - 1}{2n^2 + 5} converges, and find its limit if it exists.
  2. 2 Test the series ∑n=1∞nn3+1\sum_{n=1}^{\infty} \frac{n}{n^3 + 1} for convergence or divergence.
  3. 3 Find the radius of convergence of the power series ∑n=1∞(x−2)nn3n\sum_{n=1}^{\infty} \frac{(x-2)^n}{n3^n}.
  4. 4 Explain why the convergence of ∑n=1∞∣an∣\sum_{n=1}^{\infty} |a_n| is a stronger condition than the convergence of ∑n=1∞an\sum_{n=1}^{\infty} a_n.