Sequences, Series, and Taylor Expansions

For every ε > 0 there exists N such that n > N implies |a_n − L| < ε. Discretized version of the epsilon-limit idea.

An increasing sequence bounded above converges (to the supremum of its terms), by completeness of ℝ. Use it when you can't guess the limit.

The limit of the partial sums S_N = a_1 + ... + a_N as N → ∞. Every series statement is secretly a statement about the sequence (S_N).

No. Convergence forces a_n → 0, but not conversely: the harmonic series ∑1/n diverges though 1/n → 0 (group terms into blocks each ≥ 1/2).

∑_{n≥0} r^n = 1/(1−r) for |r| < 1; diverges otherwise. It's the only series you sum exactly and the comparison standard for everything else.

∑1/n^p converges iff p > 1. Proof: for f positive and decreasing, ∑f(n) and ∫₁^∞ f(x)dx converge together.

If lim|a_{n+1}/a_n| = L < 1 or limsup|a_n|^{1/n} = L < 1, the series converges absolutely; L > 1 diverges; L = 1 is inconclusive. They're 'eventually geometric' comparisons.

If b_n decreases monotonically to 0, then ∑(−1)^n b_n converges, and the error is bounded by the first omitted term.

∑|a_n| < ∞ implies ∑a_n converges; the converse fails, e.g. ∑(−1)^{n+1}/n = ln 2 converges conditionally.

A conditionally convergent series can be rearranged to converge to any real number or diverge; absolutely convergent series are rearrangement-proof. Infinite sums are limits, not sums.

Induction gives a_n < 2 and increasing, so it converges by MCT; passing to the limit gives L = √(2+L), L = 2. You must prove convergence BEFORE solving for L.