Ms14–Ms16 - Geometric series and sums to infinity

Ms14–Ms16 - Geometric series and sums to infinity

Use a fixed multiplier to find geometric terms and finite sums. You will justify the convergence condition for an infinite sum and distinguish a term limit from a partial-sum limit, including the awkward boundary cases.

A constant multiplier gives geometric terms

Can an endless list of alternating payments have a definite total? Consider 12−6+3−3/2+⋯12-6+3-3/2+\cdots. The totals start 12,6,9,7.5,…12,6,9,7.5,\ldots: they jump above and below 88. This lesson explains both the jumps and the limiting total.

You already know

A sequence lists values; a series adds values. The first nn terms have a partial sum SnS_n. Laws of indices combine repeated multiplication, and a negative base needs brackets when raised to a power.

The fixed multiplier gives the list its structure.

Geometric sequence

A geometric sequence, also called a geometric progression or GP, satisfies un+1=runu_{n+1}=ru_n for a constant common ratio rr. For example, 12,−6,3,…12,-6,3,\ldots has r=−1/2r=-1/2. The list 12,6,0,…12,6,0,\ldots is not a GP: no fixed multiplier gives both steps.

For nonzero consecutive terms, divide the later by the earlier to obtain rr. If a term is zero, that division is invalid. The recurrence itself still makes sense for all-zero lists and for a,0,0,…a,0,0,\ldots when r=0r=0.

Geometric term

For first term aa, ratio rr and positive integer nn,

un=arn−1u_n=ar^{n-1}

For r=0r=0, state u1=au_1=a and un=0u_n=0 for n≥2n\ge2, avoiding 000^0.

Not in the formula sheet: learn it.

Non-consecutive terms need particular care: even powers hide a sign.

Worked example 1: recover a ratio from consecutive terms

Question. A GP has u2=6u_2=6, u3=18u_3=18. Determine aa and rr.

Plan. Divide consecutive terms, then substitute into u2=aru_2=ar.

r=u3u2=186=3one stepa=63=2use ar=6\begin{aligned} r&=\frac{u_3}{u_2}=\frac{18}{6}=3 && \text{one step} \\ a&=\frac6{3}=2 && \text{use }ar=6 \end{aligned}

Why does division give rr? Consecutive nonzero terms differ by exactly one multiplication.

Answer. (a,r)=(2,3)(a,r)=(2,3).

Check. The first terms are 2,6,182,6,18, as required.

Change only which later position is supplied: keep u2=6u_2=6, but put 1818 at position four. There are now two multiplications between the known values.

Worked example 2: two possible signs

Question. A GP has u2=6u_2=6, u4=18u_4=18. Determine all possible (a,r)(a,r).

Plan. Divide the term equations to obtain r2r^2, retaining both real roots.

r2=u4u2=3two-step ratior=±3both square rootsa=6r=±23matching signs\begin{aligned} r^2&=\frac{u_4}{u_2}=3 && \text{two-step ratio} \\ r&=\pm\sqrt3 && \text{both square roots} \\ a&=\frac6r=\pm2\sqrt3 && \text{matching signs} \end{aligned}

Why are the signs matched? The product arar must be positive 66.

Answer. (a,r)=(23,3)(a,r)=(2\sqrt3,\sqrt3) or (−23,−3)(-2\sqrt3,-\sqrt3).

Check. In either case ar=6ar=6 and ar3=(ar)r2=6(3)=18ar^3=(ar)r^2=6(3)=18.

Given a pair, any later term follows from the formula; for instance, a=12,r=−1/2a=12,r=-1/2 gives u5=12(−1/2)4=3/4u_5=12(-1/2)^4=3/4. The exponent counts four changes after the first term.

Watch out

Wrong: r2=3⇒r=3r^2=3\Rightarrow r=\sqrt3 only. Right: keep both signs unless information in the question excludes one. A negative ratio alternates signs when a≠0a\ne0.

Change the later term value once more, keeping the two-step separation.

Subtract a shifted sum

To total a GP, multiply its sum by the ratio and subtract. The middle terms cancel because every shifted term matches the next unshifted one.

Sn=a+ar+⋯+arn−1rSn=ar+ar2+⋯+arn(1−r)Sn=a−arnsubtract matching terms\begin{aligned} S_n&=a+ar+\cdots+ar^{n-1} \\ rS_n&=ar+ar^2+\cdots+ar^n \\ (1-r)S_n&=a-ar^n && \text{subtract matching terms} \end{aligned}

Only the denominator restriction matters for a finite total, however large the terms become.

Finite geometric sum

For positive integer nn and r≠1r\ne1,

Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

In the formula sheet.

The equivalent form a(rn−1)/(r−1)a(r^n-1)/(r-1) follows by changing both signs. At r=1r=1, use Sn=naS_n=na directly; that boundary case is not printed as part of the fraction formula.

Worked example 3: a positive fractional ratio

Question. Find the exact first-four-term sum when a=12a=12, r=1/2r=1/2.

Plan. Use the finite formula and keep fractions exact.

S4=12{1−(1/2)4}1−1/2finite sum=24(1516)=452evaluate\begin{aligned} S_4&=\frac{12\{1-(1/2)^4\}}{1-1/2} && \text{finite sum} \\ &=24\left(\frac{15}{16}\right)=\frac{45}{2} && \text{evaluate} \end{aligned}

Why use a finite formula? Four terms are requested, so later terms contribute nothing to this answer.

Answer. S4=45/2S_4=45/2.

Check. 12+6+3+3/2=45/212+6+3+3/2=45/2.

Change only the ratio's sign. The same fourth power occurs, but the denominator changes.

Worked example 4: an alternating fractional ratio

Question. Find the exact first-four-term sum when a=12a=12, r=−1/2r=-1/2.

Plan. Evaluate the signed ratio in the denominator and its power separately.

S4=12{1−(−1/2)4}1+1/2signed ratio=8(1516)=152simplify\begin{aligned} S_4&=\frac{12\{1-(-1/2)^4\}}{1+1/2} && \text{signed ratio} \\ &=8\left(\frac{15}{16}\right)=\frac{15}{2} && \text{simplify} \end{aligned}

Why does the numerator stay the same? An even power removes the ratio's sign, while the denominator still retains it.

Answer. S4=15/2S_4=15/2.

Check. 12−6+3−3/2=15/212-6+3-3/2=15/2.

Ratios outside the convergence interval are also allowed for finite sums. For a=3,r=−2,n=4a=3,r=-2,n=4, direct addition and the formula both give 3−6+12−24=−153-6+12-24=-15.

Watch out

Wrong: reject a finite sum because ∣r∣>1|r|>1. Right: convergence is only an issue when taking infinitely many terms. For finite sums, handle r=1r=1 separately and otherwise use the formula with its signs intact.

Change the magnitude of the negative fractional ratio for your turn.

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