Infinite geometric series – partial sums approaching the sum to infinity
MathematicsSequences & Financial MathAges 16–17
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Sign in to playAdd the terms uₙ = u₁·qⁿ⁻¹ as jumps along a number line (Zeno's paradox) and watch the graph of the partial sums Sₙ. When |q| < 1, Sₙ approaches S = u₁/(1 − q); when |q| ≥ 1 the partial sums diverge. Change the common ratio q and the first term u₁, and read the gap |S − Sₙ| and how many terms are needed to close it.
Lesson: Sum to infinity of a geometric series
What it shows
A geometric series adds terms that are each q times the previous one, and its partial sum is Sₙ = u₁(1 − qⁿ)/(1 − q). When |q| < 1 the factor qⁿ shrinks towards 0, so the partial sums approach a fixed value, the sum to infinity S = u₁/(1 − q). This resolves Zeno's paradox: infinitely many ever-shorter steps can cover a finite distance. When |q| ≥ 1 the terms do not shrink, so the partial sums grow without bound or oscillate and the series diverges. The simulation adds at most 30 terms, so the limit appears as a trend.
How to use
Set First term u₁ and Common ratio q with the sliders, then press Play to watch the jumps along the number line, or Add 1 term to go one step at a time. Use the Presets buttons to compare q = 1/2, −1/2 and 9/10 with the divergent cases q = 1, −1 and 6/5. Read the gap |S − Sₙ| in the table and the number of terms needed to get within 0.001 of S.
Parameters you can change
- First term u₁ 0.1–4
- Common ratio q -1.5–1.5
- Maximum number of terms 5–30
- Show data table
Questions to explore
- With u₁ = 1 and q = 1/2, after how many terms is Sₙ within 0.001 of 2?
- Why is Sₙ sometimes above and sometimes below S when q is negative?
- Can the formula S = u₁/(1 − q) be used when q = 6/5, and why?
