Question

Explain how to tell if a geometric series is convergent or divergent.
Include an example of a convergent series and an example of a divergent series in your explanation. EASY WAY

Answer 1

Easy rule:
- A geometric series is of the form a + ar + ar^2 + ar^3 + ... = ∑_{n=0}^∞ a r^n.
- It converges exactly when |r| < 1. If it converges, the sum is S = a/(1 − r).
- If |r| ≥ 1 it diverges (terms do not go to 0 or partial sums do not approach a limit).

Why (intuitively): the nth term is ar^n. If |r|<1 then r^n → 0, and the partial sums settle to a/(1−r). If |r|≥1 the terms don't shrink to 0 (or they oscillate), so the sum cannot approach a finite limit.

Convergent example:
- Take a = 2, r = 1/3. Series: 2 + 2(1/3) + 2(1/3)^2 + ...
- |r| = 1/3 < 1, so it converges. Sum = 2/(1 − 1/3) = 2/(2/3) = 3.

Divergent example:
- Take a = 5, r = 2. Series: 5 + 10 + 20 + 40 + ...
- |r| = 2 ≥ 1, terms grow and the series diverges (no finite sum).

Quick special notes:
- r = 1 gives a + a + a + ... which diverges unless a = 0.
- r = −1 gives a − a + a − a + ... which oscillates and does not converge (unless a = 0).