Interactive visualiser

Geometric series and the sum to infinity

Choose the first term u1u_1 and the common ratio rr, then add terms one at a time. See the partial sums SnS_n close in on S∞S_\infty when ∣r∣<1|r| < 1, and fail to settle when ∣r∣≥1|r| \geq 1.

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What you are looking at

In a geometric sequence each term is the previous one multiplied by the common ratio, so un=u1rn−1u_n = u_1 r^{n-1}. Adding the first nn terms gives Sn=u1(1−rn)1−rS_n = \dfrac{u_1(1 - r^n)}{1 - r} for r≠1r \neq 1. The Bars & sums view draws each term as a green bar and joins the partial sums with a blue line, so you can see how much each new term adds. The Number line view shows the same sums as a chain of hops.

When ∣r∣<1|r| < 1, rn→0r^n \to 0 as nn grows, so SnS_n approaches S∞=u11−rS_\infty = \dfrac{u_1}{1 - r}, drawn as a dashed yellow line. The part still missing after nn terms is S∞−Sn=u1rn1−rS_\infty - S_n = \dfrac{u_1 r^n}{1 - r}, and the Table view lists it beside unu_n and SnS_n. When ∣r∣≥1|r| \geq 1 the terms do not shrink, the partial sums have no finite limit and the visualiser reports that the series diverges.

Try this

  1. 1Start from u1=8u_1 = 8, r=0.5r = 0.5 and press Play n. Each new term covers half of the remaining distance to S∞=16S_\infty = 16; after 1010 terms the gap is 161024=0.015625\dfrac{16}{1024} = 0.015625.
  2. 2Change rr to −0.5-0.5. The partial sums now land alternately above and below S∞=81.5=163S_\infty = \dfrac{8}{1.5} = \dfrac{16}{3}, closing in from both sides; the Number line view shows the hops changing direction.
  3. 3Drag rr slowly towards 11 and watch S∞=u11−rS_\infty = \dfrac{u_1}{1 - r} grow while the partial sums take longer to get near it. At r=1r = 1 every term equals u1u_1, so Sn=n u1S_n = n\,u_1.
  4. 4Hide the formulas and open the Table view. The SnS_n column shows question marks, so you can predict each partial sum before revealing it.

Mistakes students make

  • Using S∞=u11−rS_\infty = \dfrac{u_1}{1 - r} when ∣r∣≥1|r| \geq 1. The formula holds only for ∣r∣<1|r| < 1; otherwise there is no sum to infinity.
  • Writing un=u1rnu_n = u_1 r^n. The first term is u1r0u_1 r^0, so the nnth term is u1rn−1u_1 r^{n-1}.
  • Mishandling a negative ratio. With r=−0.5r = -0.5, 1−r=1.51 - r = 1.5, not 0.50.5, so S∞=u11.5S_\infty = \dfrac{u_1}{1.5}.

Where it is taught

  • IB Maths AA SL/HL 1.3 · Geometric sequences and series
  • IB Maths AA SL/HL 1.8 · Infinite convergent geometric series
  • Edexcel A-Level Maths (9MA0) · Pure Year 2 · Sequences and series: geometric series and sum to infinity
  • AP Calculus BC Unit 10 · Working with geometric series (10.2)
  • Greek Lykeio (Α΄ Λυκείου) · Algebra · Geometric progressions

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