5B - Convergent geometric series

5B - Convergent geometric series

Can infinitely many nonzero terms have a finite total? You will interpret a sum to infinity through partial sums, use the geometric convergence condition, and combine finite and infinite information to recover parameters or reach a target total.

Why an infinite sum can be finite

The sum starts 4+2+1+12+⋯4+2+1+\tfrac12+\cdots. Could its total be 88, even though no finite addition reaches 88?

The first few totals are 4,6,7,7.5,7.75,…4,6,7,7.5,7.75,\ldots. Each extra term fills half the remaining gap to 88. The gap becomes arbitrarily small as more terms are included. Calling 88 the sum to infinity describes this limiting behaviour; it does not mean there is a final term.

Convergent series

A partial sum, SnS_n, is the total of the first nn terms. A series is convergent if these partial sums approach one finite value as nn increases without bound. That value is its sum to infinity, S∞S_\infty.

For a geometric series with a nonzero first term, the decisive condition is ∣r∣<1|r|<1, equivalently −1<r<1-1<r<1. Then rnr^n approaches zero. If r<0r<0, it changes sign while its magnitude shrinks; the partial sums can alternate above and below the limiting total.

[DIAGRAM: asset_name: Partial sums approaching finite limits; asset_slug: convergent-partial-sums; caption: Positive and negative half-ratios both give convergent partial sums; description: Two discrete partial-sum plots start with first term 4. Ratio one half gives totals 4,6,7,7.5 and onwards approaching the horizontal line S equals8 from below. Ratio minus one half gives totals 4,2,3,2.5 and onwards approaching the horizontal line S equals8over3 alternately from above and below. Only integer n values are plotted.]
Positive and negative half-ratios both give convergent partial sums

The boundaries matter:

RatioBehaviour when a≠0a\ne0
r=1r=1Sn=naS_n=na, so no finite limit
r=−1r=-1Partial sums alternate between aa and 00, so no single limit
∣r∣>1\lvert r\rvert>1Terms do not approach zero, and the partial sums do not converge
∣r∣<1\lvert r\rvert<1rnr^n approaches zero and the geometric series converges

At r=0r=0, only the first term can be nonzero, so the total is simply aa. The all-zero series is a trivial exception to the nonzero-first-term discussion: its partial sums are zero whatever multiplying ratio is assigned.

Watch out

The test r<1r<1 is incomplete: −2<1-2<1, but a ratio of −2-2 makes term sizes grow. Use ∣r∣<1|r|<1, which includes shrinking negative ratios and excludes both boundary values.

Check each common ratio, including its sign.

Using the sum to infinity

The finite sum formula shows why the convergence condition produces a simple infinite total.

When ∣r∣<1|r|<1, the factor rnr^n in Sn=a(1−rn)/(1−r)S_n=a(1-r^n)/(1-r) approaches zero. The numerator therefore approaches aa, while the denominator stays fixed and nonzero.

Geometric sum to infinity

For ∣r∣<1|r|<1,

S∞=a1−rS_\infty=\frac{a}{1-r}

On the formulae sheet, together with the convergence condition.

Identify the first term, find the signed ratio, and check the condition before substituting. A value from the fraction alone is not evidence that an infinite sum exists.

Worked example 1: a positive fractional ratio

Question. Find the sum to infinity of 9+6+4+83+⋯9+6+4+\tfrac83+\cdots.

Plan. Use a=9a=9 and r=23r=\tfrac23, after checking convergence.

∣r∣=23<1the series convergesS∞=91−2/3substitute=27divide by 1/3\begin{aligned} |r|&=\frac23<1 &&\text{the series converges}\\ S_\infty&=\frac9{1-2/3} &&\text{substitute}\\ &=27 &&\text{divide by }1/3 \end{aligned}

Answer. 2727.

Check. Removing the first 99 from a total of 2727 leaves 1818. The remaining series is the original multiplied by 23\tfrac23, and 23(27)=18\tfrac23(27)=18, as required.

Change only the sign of the ratio. Subtracting a negative ratio makes the denominator larger than 11.

Worked example 2: an alternating convergent series

Question. Find the sum to infinity of 9−6+4−83+⋯9-6+4-\tfrac83+\cdots.

Plan. Check the magnitude of r=−23r=-\tfrac23, then retain its sign in the denominator.

∣r∣=23<1negative ratios can convergeS∞=91−(−2/3)substitute the signed ratio=275divide by 5/3\begin{aligned} |r|&=\frac23<1 &&\text{negative ratios can converge}\\ S_\infty&=\frac9{1-(-2/3)} &&\text{substitute the signed ratio}\\ &=\frac{27}{5} &&\text{divide by }5/3 \end{aligned}

Answer. 275\tfrac{27}{5}.

Check. Removing the first term leaves 275−9=−185\tfrac{27}{5}-9=-\tfrac{18}{5}, exactly −23-\tfrac23 times 275\tfrac{27}{5}. The negative remainder is consistent with the next term being negative.

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