Active booklet

HSC Sequences

Home

HSC Sequences · Chapter guide

Infinite Series & Recurring Decimals

When |r|<1, a geometric series sums to a/(1-r). Recurring decimals are infinite GPs in disguise — split terminating and repeating parts, then convert exactly to a fraction.

Learning outcomes

  • State the |r|<1 condition for an infinite GP
  • Evaluate S_∞ = a/(1-r)
  • Convert pure and mixed recurring decimals to fractions
  • Spot divergence when |r|≥1

When does an infinite GP have a sum?

A finite GP always has a sum. An infinite geometric series a+ar+ar2+a+ar+ar^2+\cdots converges only when r<1|r|<1. In that case

S=a1r.S_\infty = \frac{a}{1-r}.

If r1|r|\ge 1, the terms do not tend to zero (or fail to shrink fast enough), and there is no finite limiting sum. HSC markers expect you to check r<1|r|<1 before writing SS_\infty.

Worked example 1 — Convergence check

Does n=03(23)n\sum_{n=0}^{\infty} 3\left(-\frac23\right)^n converge? If so, find its sum.

Here a=3a=3 and r=23r=-\frac23. Since 23<1\left|-\frac23\right|<1, the series converges:

S=31(23)=353=95.S_\infty=\frac{3}{1-(-\frac23)}=\frac{3}{\frac53}=\frac95.

If the ratio had been 32-\frac32, you would stop: r>1|r|>1, no SS_\infty.

Recurring decimals as geometric series

A pure recurring decimal such as 0.2˙=0.2220.\dot{2}=0.222\ldots is the series

210+2100+21000+,\frac{2}{10}+\frac{2}{100}+\frac{2}{1000}+\cdots,

an infinite GP with first term 210\frac{2}{10} and ratio 110\frac{1}{10}. Mixed recurring decimals (e.g. 0.12˙0.1\dot{2}) split into a terminating place-value part plus a pure recurring tail.

Worked example 2 — Pure and mixed recurring

Convert 0.20.2, 0.2˙0.\dot{2}, and 0.12˙0.1\dot{2} to fractions.

  • Terminating: 0.2=210=150.2=\frac{2}{10}=\frac15.
  • Pure recurring: 0.2˙0.\dot{2} has a=210a=\frac{2}{10}, r=110r=\frac{1}{10}, soS=2/1011/10=29.S=\frac{2/10}{1-1/10}=\frac29.
  • Mixed: 0.12˙=0.1+0.02˙0.1\dot{2}=0.1+0.0\dot{2}. The tail 0.02˙0.0\dot{2} has a=2100a=\frac{2}{100}, r=110r=\frac{1}{10}:2/10011/10=290=145.\frac{2/100}{1-1/10}=\frac{2}{90}=\frac{1}{45}. Adding 110=990\frac{1}{10}=\frac{9}{90} gives 1190\frac{11}{90}.

Worked example 3 — Another recurring form

Express 0.420.\overline{4}2 ideas carefully: identify which digits repeat, write the first repeating block as aa, and take r=10kr=10^{-k} where kk is the repeat length. The same SS_\infty formula applies; only aa and rr change.

For a single repeating digit after the point, denominators of 99, 9999, 999999 appear naturally from 1r=110k1-r=1-10^{-k}.

Exam strategy

  1. Write the decimal as a series (or as terminating + series).
  2. State aa and rr explicitly.
  3. Verify r<1|r|<1 (always true for standard decimal expansions with r=10kr=10^{-k}).
  4. Simplify the fraction fully.

Common mistakes

  • Writing SS_\infty without checking r<1|r|<1 on abstract GPs.
  • Treating a mixed recurring decimal as if every digit repeats.
  • Off-by-one errors in the first term’s place value (/10/10 vs /100/100).
  • Leaving unsimplified fractions when a clear reduction exists.

Practice (hints only)

  1. Convert 0.6˙0.\dot{6} and 0.27˙0.2\dot{7} to fractions. (GP with r=110r=\frac1{10}.)
  2. For which rr does 2+2r+2r2+2+2r+2r^2+\cdots converge, and what is SS_\infty? (Need r<1|r|<1.)

Where next

With infinite sums under control, study telescoping sums, sigma shifts, and induction — techniques that go beyond a single closed GP formula.

Open section in PDF (p. 8)Full study guideMore on the blog

Related worked solutions