1.04i, 1.04j - Geometric series and sums to infinity | OCR A-Level Mathematics A

Find geometric terms and finite totals by tracking a constant multiplier. You will derive the sum formula, decide when an infinite sum exists and separate the behaviour of terms from the behaviour of their totals.

A constant multiplier gives geometric terms

Can you keep adding positive numbers forever without the total exceeding $16$ ? Start with $8+4+2+1+\cdots$ . Each new amount is half its predecessor. The totals are $8,12,14,15,\ldots$ : they move towards $16$ , while the added terms move towards zero. We will make that claim precise.

You already know

A sequence lists terms; a series adds them. A recurrence describes one term using previous terms. Index laws give $r^p r^q=r^{p+q}$ , including negative $r$ with integer powers.

Equal multiplication needs a different term formula from equal addition.

Geometric sequence

A geometric sequence, or geometric progression (GP), satisfies $u_{n+1}=ru_n$ for a fixed common ratio $r$ . With first term $a$ , it begins $a,ar,ar^2,\ldots$ . For example, $3,-6,12,\ldots$ has $r=-2$ . The list $3,6,9,\ldots$ is not geometric.

For nonzero consecutive terms, $r=u_{n+1}/u_n$ . Do not divide by zero. The recurrence still defines the all-zero case and $a,0,0,\ldots$ when $r=0$ .

Geometric term

For positive integer $n$ ,

u_n=ar^{n-1}

For $r=0$ , use $u_1=a$ and $u_n=0$ for $n\ge2$ , avoiding $0^0$ .

Not in the formula sheet: learn it.

The exponent again counts changes after the first term.

Worked example 1: a positive multiplier

Question. Find the sixth term of the GP with $a=3$ , $r=2$ .

Plan. Multiply the first term by the fifth power of the ratio.

\begin{aligned} u_6&=3(2)^5 && \text{five multiplications} \\ &=96 && \text{evaluate} \end{aligned}

Why the fifth power? Position six is five steps after position one.

Answer. $u_6=96$ .

Check. The terms are $3,6,12,24,48,96$ .

Change only the ratio's sign. The magnitudes stay the same, but the signs alternate.

Worked example 2: a negative multiplier

Question. Find the sixth term when $a=3$ , $r=-2$ .

Plan. Keep the negative ratio in brackets before raising it to an odd power.

\begin{aligned} u_6&=3(-2)^5 && \text{term formula} \\ &=3(-32)=-96 && \text{odd power is negative} \end{aligned}

Why keep brackets? The whole ratio, including its sign, is multiplied five times.

Answer. $u_6=-96$ .

Check. $u_5=48$ , and $48(-2)=-96$ .

Separated terms may leave two possible ratios. If $u_2=6$ and $u_4=24$ , dividing gives $r^2=4$ . Both $(a,r)=(3,2)$ and $(-3,-2)$ fit. Consecutive nonzero terms determine the sign, but a square need not.

Watch out

Wrong: $r^2=4\Rightarrow r=2$ only. Right: $r=\pm2$ , unless the question supplies a sign restriction. Substitute each ratio to find its matching first term.

Change the magnitude of the negative ratio for your turn.

Subtract a shifted sum

Multiplying a geometric sum by its ratio aligns almost every term with another term. Subtraction then leaves just the two ends.

\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 the rows} \end{aligned}

There is no convergence condition for a finite sum. The only denominator restriction is $r\ne1$ .

Finite geometric sum

For positive integer $n$ and $r\ne1$ ,

S_n=\frac{a(1-r^n)}{1-r}

In the formula sheet.

Multiplying numerator and denominator by $-1$ gives the equivalent form $a(r^n-1)/(r-1)$ , often convenient for $r>1$ . At $r=1$ , sum the constant terms directly: $S_n=na$ . That separate case is not printed as a geometric-sum formula.

Worked example 3: a growing finite sum

Question. Find the sum of the first five terms when $a=5$ , $r=2$ .

Plan. Substitute into the finite formula, using the version with positive denominator.

\begin{aligned} S_5&=\frac{5(2^5-1)}{2-1} && \text{finite sum} \\ &=5(32-1)=155 && \text{evaluate} \end{aligned}

Why is $r=2$ allowed? Only five terms are being added; no infinite limit is involved.

Answer. $S_5=155$ .

Check. $5+10+20+40+80=155$ .

Change only the ratio's sign. Cancellation now makes the total much smaller.

Worked example 4: an alternating finite sum

Question. Find the first-five-term sum when $a=5$ , $r=-2$ .

Plan. Retain the signed ratio in both its power and the denominator.

\begin{aligned} S_5&=\frac{5\{1-(-2)^5\}}{1-(-2)} && \text{finite formula} \\ &=\frac{5(33)}{3}=55 && \text{evaluate signs} \end{aligned}

Why is the denominator $3$ ? Subtracting a negative ratio adds $2$ .

Answer. $S_5=55$ .

Check. $5-10+20-40+80=55$ .

The total depends on how many positive and negative terms are included. It is not simply the positive-ratio total with a minus sign.

Watch out

Wrong: $1-(-2)= -1$ . Right: $1-(-2)=3$ . Bracket a negative ratio everywhere it occurs; evaluate its power before subtracting.

Change the negative ratio's magnitude again.

A convergent series has a limiting sum

An infinite sum describes where the partial sums approach, rather than a final addition that finishes the list. For $|r|<1$ , the power $r^n$ tends to zero, so the finite formula has a finite limit.

Convergent series

A series converges if its partial sums approach a finite limit. That limit is its sum to infinity, $S_\infty$ . If there is no finite limit, the series diverges.

For a nonzero first term, a geometric series converges exactly when $-1<r<1$ . Modulus writes both bounds compactly.

Geometric sum to infinity

When $|r|<1$ ,

S_\infty=\frac{a}{1-r}

In the formula sheet.

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1.04i, 1.04j - Geometric series and sums to infinity

A constant multiplier gives geometric terms

Subtract a shifted sum

A convergent series has a limiting sum

Continue this lesson.

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