MATH1961 · Mathematics 1a (advanced)
Series
A series is the limit of its partial sums, and the whole theory rests on one question — do the terms bunch toward a single number as n grows? The Advanced course wants the rigorous answer (the ε–N definition of sequence convergence), not just “it looks like it goes to 2”. You start with sequences and the ε–N definition (read in order: for-all ε, then there-exists N), the monotone-plus-bounded convergence theorem (the completeness axiom again), and proving divergence by two subsequences with different limits. The geometric series is the prototype every power series imitates — it gives the radius of convergence — and the centrepiece is Taylor’s theorem with the Lagrange remainder, proved via the Cauchy MVT, plus the lemma that makes the series actually represent ex, sin and cos.
What this chapter covers
- 01Sequences and their limits — the ε–N definition, made usable
- 02The toolkit: algebra of limits, squeeze, monotone + bounded ⇒ convergent
- 03Proving divergence by two subsequences
- 04Geometric series, power series & the radius of convergence
- 05Taylor’s theorem with the Lagrange remainder, and why the series represents ex, sin, cos
Worked example: sum a geometric series and state when it converges
- +1Write the partial sum in closed form. For x ≠ 1, sn = 1 + x + … + xn = (1 − xn+1)/(1 − x). The series is the limit of sn.
- +1Decide convergence by the remainder xn+1. If |x| < 1 then xn+1 → 0, so sn → 1/(1 − x). If |x| ≥ 1 the terms do not even tend to 0, so the series diverges — the boundary is sharp.
- +1(a) Conclude. The series converges exactly for |x| < 1, with sum 1/(1 − x).
- +1(b) Apply at x = 1/3. Since |1/3| < 1, the sum is 1/(1 − 1/3) = 1/(2/3) = 3/2.
Key terms
- Sequence convergence (ε–N)
- an → L means: for every ε > 0 there is an N such that |an − L| < ε for all n > N. The order for-all ε then there-exists N is the content — N may depend on ε, and a smaller ε usually needs a larger N.
- Monotone convergence theorem
- A sequence that is monotone (always increasing or always decreasing) and bounded must converge. It is a direct consequence of the completeness of ℝ (the bound supplies a sup or inf that the sequence climbs to), and it is the standard way to prove a recursively-defined sequence converges.
- Geometric series
- Σk≥0 xk, which converges iff |x| < 1, with sum 1/(1 − x). It is the one series you can sum in closed form and the prototype every power series imitates; the cut |x| < 1 is exactly where the remainder xn+1 → 0.
- Radius of convergence
- For a power series Σ ck(x − a)k, the number R such that the series converges for |x − a| < R and diverges for |x − a| > R. It is found by the ratio or root test; behaviour exactly at |x − a| = R must be checked separately.
- Taylor's theorem (Lagrange remainder)
- f equals its degree-n Taylor polynomial plus a remainder Rn = f(n+1)(c)/(n+1)! · (x − a)n+1 for some c between a and x. Proved via the Cauchy MVT, it bounds the approximation error — and showing Rn → 0 is what makes the Taylor series actually equal ex, sin x and cos x.
Series FAQ
How do I prove a sequence converges in the Advanced unit?
Either directly from the ε–N definition (given ε, find an N past which |an − L| < ε), or — especially for recursively-defined sequences — by showing it is monotone and bounded and invoking the monotone convergence theorem, which is the completeness axiom in disguise. ‘It looks like it settles down’ earns nothing; the marks are in the rigorous argument.
How do I prove a sequence diverges?
Exhibit two subsequences with different limits. For an = (−1)n, the even terms → 1 and the odd terms → −1, so no single limit L can exist. As with limits of functions, one subsequence is never enough — that shortcut is a recurring slip.
Where does the radius of convergence come from?
From treating a power series like a geometric series: apply the ratio (or root) test to |ck(x − a)k|, and the condition that the limiting ratio is < 1 gives |x − a| < R. Inside R the series converges; outside it diverges; and the two endpoints must be tested by hand because the ratio test is inconclusive there.
Why does Taylor's theorem need the remainder?
Because the Taylor polynomial is only an approximation; the series equals the function only if the error vanishes. The Lagrange remainder Rn = f(n+1)(c)/(n+1)! · (x − a)n+1 bounds that error, and the lemma |x|n+1/(n+1)! → 0 shows Rn → 0 for ex, sin and cos — which is precisely why their Taylor series represent them everywhere.
Exam move
Treat sequence convergence as the foundation: be able to write the ε–N definition in order and prove a simple limit from it, and keep the monotone-plus-bounded theorem ready for recursive sequences. Memorise the geometric series result (converges iff |x| < 1, sum 1/(1 − x)) because it is the template for every power series and the radius of convergence. For Taylor’s theorem, learn the statement with the Lagrange remainder and the Cauchy-MVT proof idea, and practise the remainder-bound argument that shows Rn → 0 — that is the Advanced premium over just quoting the Maclaurin series. Rehearse the two-subsequence divergence template alongside.
