← Unit 10

Unit 10 Cheat Sheet

Every convergence test, every must-know series, and the one thing the unit really tests: which tool to reach for. The tests look interchangeable until you know their tells — so the big block below is a decision guide, not just a list.
→ Test yourself

Series Foundations the basics

A series converges if its partial sums approach a finite limit.

geometric series
p-series

Two you just know: a geometric series converges exactly when \(|r|\lt 1\) (and you can find its sum); a p-series converges exactly when \(p\gt 1\). The harmonic series \((p=1)\) diverges.

trap
  • Geometric sum \(\tfrac{a}{1-r}\) uses \(a=\) the first term actually present.

The Divergence Test always first

The one quick check to run before anything else.

nth-term (divergence) test

If the terms don't shrink to zero, the series can't converge. But it only ever shows divergence — if \(\lim a_n=0\), the test is inconclusive and you need a real test.

trap
  • \(\lim a_n=0\) does not prove convergence. The harmonic series has \(a_n\to0\) yet diverges.

Error Bounds how close?

How far a partial sum / Taylor polynomial is from the true value.

Lagrange error (Taylor)
alternating series error
Which error bound?
Taylor polynomial→ Lagrange: \(M=\max|f^{(n+1)}|\) on the interval
alternating series→ error ≤ the first term you dropped
trap
  • Lagrange uses the \((n{+}1)\)th derivative and \((n{+}1)!\), not the \(n\)th.
  • \(M\) is the max of the derivative on the whole interval, not its value at a point.

The Convergence Tests the toolkit

Each test has a tell — a series shape it's built for.

Geometric
Form \(ar^n\). Converges iff \(|r|\lt 1\).
p-series
Form \(\tfrac{1}{n^p}\). Converges iff \(p\gt 1\).
Ratio
Factorials or \(n\)th powers. \(L=\lim\big|\tfrac{a_{n+1}}{a_n}\big|\): \(L\lt 1\) conv, \(L\gt 1\) div, \(L=1\) try another.
Root
Whole term to the \(n\)th power. \(L=\lim\sqrt[n]{|a_n|}\), same thresholds.
Integral
\(a_n=f(n)\), \(f\) positive, decreasing, integrable. Series and \(\int_1^\infty f\) do the same thing.
Comparison
Bound by a known series: smaller-than-convergent conv; bigger-than-divergent div.
Limit Comp.
\(\lim\tfrac{a_n}{b_n}=c\gt 0\): both do the same. Pick \(b_n\) by dominant terms.
Alternating
\((-1)^n a_n\) with \(a_n\) decreasing to 0 → converges.
Which test? — read the shape
terms ↛ 0→ nth-term: diverges, done
\(ar^n\) / \(1/n^p\)→ geometric / p-series by inspection
factorials, \(n\)th powers→ Ratio (or Root)
looks like a p-series→ Limit Comparison to \(1/n^p\)
nice to integrate→ Integral Test
\((-1)^n\) out front→ Alternating Series Test
trap
  • Ratio/Root with \(L=1\) tells you nothing — switch tests.
  • Absolute vs conditional: if \(\sum|a_n|\) converges it's absolute; if only \(\sum a_n\) does, conditional.

Absolute vs. Conditional which kind?

When a series has negative terms, there are two grades of convergence.

Test the absolute value first
\(\sum|a_n|\) converges→ absolutely convergent (the strong kind)
\(\sum|a_n|\) divergesbut \(\sum a_n\) converges → conditionally convergent
\(\sum a_n\) diverges→ just divergent

Why it matters: absolute convergence is sturdy — you can rearrange terms freely. Conditional convergence is fragile. The classic split: \(\sum\tfrac{(-1)^n}{n}\) is conditional (the harmonic series underneath diverges), while \(\sum\tfrac{(-1)^n}{n^2}\) is absolute.

Quick routine: strip the signs and test \(\sum|a_n|\) with whatever fits (often Ratio or a p-series comparison). Converges → absolute, and you're done. Diverges → fall back to the Alternating Series Test on \(\sum a_n\); if that converges, it's conditional.

trap
  • Run the Alternating Series Test on \(\sum a_n\), but test convergence on \(\sum|a_n|\) to decide the kind.
  • "Converges" on the AP usually means conditionally or absolutely — name which when asked.

Taylor & Maclaurin Series build the polynomial

Rebuild a function as an infinite polynomial from its derivatives at a point.

Taylor series about x = c

Maclaurin is just Taylor with \(c=0\). The \(n\)th-degree Taylor polynomial stops the sum at \(n\).

trap
  • Divide by \(n!\) — the derivatives alone aren't the coefficients.
  • "About \(x=c\)" means powers of \((x-c)\), not \(x\).

Memorize these four — they're the starting point for most problems (sub, differentiate, or integrate to get others):

e^x · all reals
sin x · all reals
cos x · all reals
geometric · |x| < 1

Radius & Interval of Convergence where it's valid

A power series only equals its function for \(x\) close enough to the center.

radius (via ratio test)
interval (before endpoints)
ln(1 + x) · −1 < x ≤ 1
Finding the interval
1. ratio testrun it on the power series, set \(L\lt 1\)
2. solveget \(|x-c|\lt R\) → open interval
3. test endsplug each endpoint in separately
4. write ituse [ ] or ( ) per endpoint result
trap
  • Always check both endpoints by substitution — the ratio test goes silent there \((L=1)\).
  • \(R=\infty\) (e.g. \(e^x,\sin,\cos\)) means all reals; \(R=0\) means only at \(x=c\).
Σ

The whole unit answers two questions: does this infinite sum settle to a number (convergence tests), and can I rebuild a function as an infinite polynomial (Taylor series)? Match the series' shape to the right test, and know your four Maclaurin series cold — everything else is built from those.

Test yourself on the exam set →
Foundations Convergence Taylor Interval Error