Check convergence before summing to infinity

Compare limits of arithmetic and geometric sums

The comparison table uses five examples. Its reasoning is about the whole partial sum, including the signs and the conditions on the parameters:

• For $\sum_{j=1}^{n}2^j=2(2^n-1)$, the power $2^n\to+\infty$, so the partial sums tend to positive infinity. The series diverges.
• Given $\sum_{j=1}^{n}1/[j(j+2)]=3/4-1/[2(n+1)]-1/[2(n+2)]$, both reciprocal remainders tend to zero. The series converges with sum $3/4$.
• For an AP, expand the partial sum:

$$S_n=\frac n2[2a+(n-1)d] =\frac d2n^2+\left(a-\frac d2\right)n.$$

For $d>0$, the quadratic term tends to positive infinity and dominates the linear term, so $S_n\to+\infty$. For $d<0$, it dominates with the opposite sign and $S_n\to-\infty$. Both cases diverge.

• For a GP with $r>1$, the source uses

$$S_n=\frac{a(r^n-1)}{r-1},\qquad r^n\to+\infty.$$

The source prints $S_n\to+\infty$. This requires $a>0$; if $a<0$, then $S_n\to-\infty$. Every such nonzero GP diverges. Keep the sign of $a$ when describing the direction of divergence.

• For a GP with $|r|<1$, the powers tend to zero, including when a negative ratio makes the terms alternate:

$$r^n\to0,\qquad S_n=\frac{a(1-r^n)}{1-r}\to\frac{a}{1-r}.$$

This series converges with sum to infinity $a/(1-r)$. Check the ratio condition before applying that formula.

The table’s callout asks about the cases $d=0$ in an AP and $r=1$ in a GP. In both cases every term equals $a$ and $S_n=na$. If $a>0$, the partial sums tend to positive infinity; if $a<0$, they tend to negative infinity. If $a=0$, the all-zero series converges to zero.

Under the source’s convention a GP has nonzero terms, so the zero case is a useful extension, not a nonzero GP.

At $r=1$, use $S_n=na$ rather than substituting into a quotient with denominator zero. The callout has no printed solution; these constant-case explanations are authored working.

Visualise a halving series and distinguish its two limits

The source’s Series 1 is

$$S_n=1+\frac12+\frac14+\frac18+\frac1{16}+\frac1{32}+\cdots+\left(\frac12\right)^{n-1}, \qquad u_n=\left(\frac12\right)^{n-1}.$$

The displayed list illustrates sufficiently many terms; for a small $n$, stop at its actual nth term. As $n$ becomes large, the next term becomes small and changes the accumulated sum by less. The Student copy says the term becomes large here; the Teacher correctly refers to the index $n$ becoming large.

The source’s rectangle has total area 2. Its left half has area 1. The right half is successively divided into regions labelled $1/2,1/4,1/8,1/16,1/32,1/64,\ldots$: each new region halves the remaining unfilled area. This describes the mathematical content of the partition diagram, including the ellipsis indicating continuing subdivisions.

After $n$ regions the unfilled area is $2(1/2)^n$, which tends to zero.

Thus the partial sums approach the full area 2. The finite GP formula confirms the picture:

$$S_n=\frac{1-(1/2)^n}{1-1/2} =2\left[1-\left(\frac12\right)^n\right]\to2.$$

The two-part notation question asks for two different limits:

$$\text{(a)}\quad\lim_{n\to\infty}u_n =\lim_{n\to\infty}\left(\frac12\right)^{n-1}=0, \qquad \text{(b)}\quad\lim_{n\to\infty}S_n=2.$$

Part (a) is the limit of a term sequence; part (b) is the limit of the partial sums, hence the sum to infinity of the series. Terms tending to zero alone do not establish convergence of a series. Here convergence follows from the partial-sum formula or the shrinking remainder area.

Check the ratio before using a sum to infinity

Series 2 contrasts with the halving series: $1+2+4+8+16+32+\cdots$ has $u_n=2^{n-1}$ and $S_n=2^n-1$. Its terms and partial sums grow without bound, so it has no finite sum to infinity.

For a nonzero GP with $r\ne1$, start from the finite formula and inspect the power:

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

If $|r|<1$, equivalently $-1<r<1$, then $r^n\to0$. This gives

$$S_\infty=\frac{a}{1-r}\qquad\text{provided }|r|<1.$$

If $r>1$, the powers grow positively and the nonzero partial sums diverge with sign determined by $a$.

If $r<-1$, the powers alternate with growing magnitude, so the partial sums do not approach any single finite value.

The source’s statement that $r^n\to+\infty$ for all $|r|>1$ needs this negative-ratio qualification; its conclusion of divergence remains correct.

At $r=-1$, partial sums alternate between $a$ and zero.

At $r=1$, $S_n=na$. These boundary cases diverge when $a\ne0$. The all-zero extension is convergent.

The source remarks give $0.1+0.01+0.001+0.0001+\cdots$ as a convergent example: $|r|=0.1<1$ and $S_\infty=0.1/(1-0.1)=1/9$. In contrast, the doubling series has $|r|=2>1$, so the infinite-sum formula is unavailable.

The three source plots show partial sums, with first term $a=3$ in every case. Horizontal position is the positive integer $n$; vertical position is $S_n$. The following exact calculated samples give a text equivalent of the plotted patterns:

For $r=2$, the plotted points climb increasingly quickly and the series diverges.

For $r=1/2$, they rise toward the horizontal line $S=6$ from below.

For $r=-1/2$, odd partial sums are above $S=2$ and even partial sums below it, with shrinking deviations.

Both shrinking-ratio series converge, although only one has monotone partial sums.

The source joins points for visual effect; a series still uses integer indices rather than a continuously varying term count. The table supplies calculated values, not measurements read off the sketches.

Explain the arithmetic exception and practise ratio tests

The source asks whether an infinite arithmetic series has a sum to infinity. Its Teacher solution expands

$$S_n=\frac n2[2a+(n-1)d] =\frac d2n^2+\left(a-\frac d2\right)n$$

and concludes divergence. For $d\ne0$ this is a quadratic polynomial and its leading term determines the direction of divergence. For $d=0$ it is instead linear: $S_n=na$. Thus every nonzero arithmetic series diverges, while the all-zero AP ($a=d=0$) converges to zero. This is the exception to the source summary table’s unqualified “N.A.” entry for arithmetic sums to infinity.

Self-Practice 8 asks for the sum to infinity if it exists in each case:

• $3,9,27,81,\ldots$ has $a=3,r=3$. Since $|r|>1$, the sum to infinity does not exist.
• $0.9,0.9^2,0.9^3,0.9^4,\ldots$ has $a=0.9,r=0.9$. It converges to $0.9/(1-0.9)=9$. The first term is not 1.
• $2,-4/3,8/9,-16/27,\ldots$ has $a=2,r=-2/3$. It converges to $2/[1-(-2/3)]=6/5$. Keep the negative sign in the sum formula; use its modulus only in the convergence test.

These are the source’s three printed answers. The accompanying ratio substitutions explain why each is valid. The comparison with an AP is about partial sums: convergence of individual terms is a different question.

Recover a GP from a term and its infinite sum

Example 27 gives $u_2=5$ and $S_\infty=20$, then asks for the first-eight-term sum. Translate the two known quantities before solving:

$$ar=5\quad\text{(1)},\qquad\frac{a}{1-r}=20\quad\text{(2)}.$$

Dividing (1) by (2) eliminates $a$:

$$\begin{aligned} r(1-r)&=\frac14,\\ r^2-r+\frac14&=0,\\ \left(r-\frac12\right)^2&=0,\\ r&=\frac12. \end{aligned}$$

This ratio meets the convergence condition. Substituting in (1) gives $a=5(2)=10$. The requested sum is finite:

$$S_8=\frac{10[1-(1/2)^8]}{1-1/2}=\frac{1275}{64}.$$

Do not return the already supplied infinite sum instead of the sum of eight terms. The source derivation starts on page 41 and continues on page 42; both belong to this exercise.

Relate a prefix to the remaining infinite tail

Example 28 gives a nonzero first term and $r=1/\sqrt[6]{2}=2^{-1/6}$. The first $n$ terms sum to all the remaining terms, so the tail excludes those first $n$ terms:

$$S_n=S_\infty-S_n\quad\Longrightarrow\quad S_\infty=2S_n.$$

Since $0<r<1$, both formulas are valid. Substitute them and cancel the nonzero common factor:

$$\begin{aligned} \frac{a}{1-2^{-1/6}} &=2\frac{a[1-(2^{-1/6})^n]}{1-2^{-1/6}},\\ 1&=2[1-(2^{-1/6})^n],\\ (2^{-1/6})^n&=\frac12,\\ 2^{n/6}&=2^1,\\ \frac n6&=1,\\ n&=6. \end{aligned}$$

The nonzero-first-term condition matters: if all terms were zero, the equality would hold for every $n$. For a general convergent nonzero GP, equality of prefix and tail reduces to $r^n=1/2$; only then use the particular ratio to find the count.