Infinite Geometric Series: Free Response

Divide a term by the one directly before it, and check the quotient against a second pair before trusting it.

$\text{(i) } \frac{12}{18} = \frac23, \qquad \frac{8}{12} = \frac23$

The ratio holds steady at $\tfrac23$, and $\left|\tfrac23\right| < 1$, so series (i) converges.

$\text{(ii) } \frac{15/2}{5} = \frac32, \qquad \frac{45/4}{15/2} = \frac32$

Here $\left|\tfrac32\right| > 1$, so series (ii) diverges.

The terms of (iii) alternate in sign with the same size every time:

$\text{(iii) } r = \frac{9}{-9} = -1.$

Since $|r| = 1$, series (iii) diverges as well; a ratio of size exactly $1$ never converges.

$\text{(iv) } \frac{6}{36} = \frac16, \qquad \frac{1}{6} = \frac16$

With $\left|\tfrac16\right| < 1$, series (iv) converges. So two of the four series pass the test that a nonzero first term and $|r|<1$ demand, one fails it by growing too large, and one fails it by landing exactly on the boundary.

Only the two convergent series from part A get a sum; the formula has nothing to say about the other two.

For (i), $a = 18$ and $r = \tfrac23$:

$S = \frac{18}{1 - \tfrac23} = \frac{18}{\tfrac13} = 54.$

For (iv), $a = 36$ and $r = \tfrac16$:

$S = \frac{36}{1 - \tfrac16} = \frac{36}{\tfrac56} = 36 \cdot \frac65 = \frac{216}{5}.$

Both sums are finite numbers the partial sums close in on: $54$ for (i) and $\tfrac{216}{5} = 43.2$ for (iv).

List a few partial sums of each divergent series and watch what they do.

For the series with $a=5$ and $r=\tfrac32$:

$S_1 = 5, \quad S_2 = 12.5, \quad S_3 = 23.75, \quad S_4 = 40.625, \ldots$

Each term is $1.5$ times the one before it, and multiplying by a number bigger than $1$ in size over and over drives the total past any number you name. That is what "diverges" means here: the partial sums pass every bound.

For the series with $a = -9$ and $r = -1$:

$S_1 = -9, \quad S_2 = 0, \quad S_3 = -9, \quad S_4 = 0, \ldots$

Nothing here grows. Every partial sum equals $-9$ or $0$, so the list stays inside a bounded pen forever, yet it never settles on a single value, and the definition of a sum requires exactly one value. So this series diverges for a completely different reason than the other one: one fails by running away, while the other fails by refusing to settle even though it never leaves a small range. Diverging does not mean blowing up; it means not closing in on one number, and a ratio of $-1$ is exactly the case that shows the two are not the same thing.

Every two-digit block of the repeating pattern is worth $\tfrac{1}{100}$ of the block before it, since it sits two decimal places further right, so read the decimal as

$0.\overline{36} = \frac{36}{100} + \frac{36}{10{,}000} + \frac{36}{1{,}000{,}000} + \cdots,$

a geometric series with $a = \tfrac{36}{100}$ and $r = \tfrac{1}{100}$. Since $\left|\tfrac{1}{100}\right| < 1$ it converges, and

$S = \frac{\tfrac{36}{100}}{1 - \tfrac{1}{100}} = \frac{\tfrac{36}{100}}{\tfrac{99}{100}} = \frac{36}{99} = \frac{4}{11}.$

The hundreds cancel, and $36$ and $99$ share the factor $9$, leaving $\tfrac{4}{11}$.

Separate the digit that does not repeat from the block that does:

$0.2\overline{7} = \frac{2}{10} + \left(\frac{7}{100} + \frac{7}{1000} + \frac{7}{10{,}000} + \cdots\right).$

The bracket is geometric with $a = \tfrac{7}{100}$ and $r = \tfrac{1}{10}$, so it sums to

$\frac{\tfrac{7}{100}}{1 - \tfrac{1}{10}} = \frac{\tfrac{7}{100}}{\tfrac{9}{10}} = \frac{7}{100}\cdot\frac{10}{9} = \frac{7}{90}.$

Add back the stray tenth, over a common denominator of $90$:

$0.2\overline{7} = \frac{2}{10} + \frac{7}{90} = \frac{18}{90} + \frac{7}{90} = \frac{25}{90} = \frac{5}{18}.$

Dividing $5$ by $18$ does return $0.27777\ldots$, which checks the answer.

The two decimals look alike, but only one of them is genuinely an infinite list of terms.

$0.36$ means $\tfrac{3}{10} + \tfrac{6}{100}$ and nothing more; every digit after the hundredths place is a $0$ that contributes nothing:

$0.36 = \frac{3}{10} + \frac{6}{100} + 0 + 0 + \cdots \quad\text{(only two nonzero terms)}.$

Adding those two fractions is a single, finite computation, the same kind of addition that worked long before this lesson existed. This decimal's own partial sums are, trivially, $0.3$, then $0.36$, then $0.36$, then $0.36$ forever: the sequence is already constant after two terms, so nothing is left for the definition of an infinite sum to determine, unlike a genuinely never-ending list of nonzero terms.

$0.\overline{36}$, by contrast, really does carry infinitely many nonzero terms, $\tfrac{36}{100} + \tfrac{36}{10{,}000} + \cdots$ forever, and no finite stopping point ever reaches the exact value: every partial sum, $0.36$, $0.3636$, $0.363636$, and so on, falls short by a smaller and smaller amount but never reaches $\tfrac{4}{11}$ exactly. Turning that into a fraction is not ordinary finite arithmetic; it requires deciding what the endless list of partial sums is closing in on, which is exactly the definition this lesson opened with.

So the classmate is mistaken. A terminating decimal is a finite sum in disguise, and finite sums needed no new definition. It is specifically the never-ending repetition that calls for one.

Divide a term by the one before it to find the ratio:

$r = \frac{32.4}{54} = 0.6, \qquad \frac{19.44}{32.4} = 0.6.$

The ratio holds steady at $0.6$, and $|0.6| < 1$, so the series converges with $a = 54$:

$S = \frac{54}{1 - 0.6} = \frac{54}{0.4} = 135.$

The error left after $n$ terms of a convergent geometric series is exactly $S\,r^{\,n}$, so the demand "within $0.05$" is the inequality

$135(0.6)^{\,n} < 0.05 \qquad\Longleftrightarrow\qquad (0.6)^{\,n} < \frac{0.05}{135} \approx 0.00037037.$

Take logarithms of both sides, which is legal because both sides are positive and $\log$ is increasing:

$n \log(0.6) < \log(0.00037037).$

Since $\log(0.6) \approx -0.2218$ is negative, dividing by it reverses the inequality:

$n > \frac{\log(0.00037037)}{\log(0.6)} \approx \frac{-3.4314}{-0.2218} \approx 15.47.$

The smallest whole number above $15.47$ is $16$. Checking the boundary is worth doing: $135(0.6)^{15} \approx 0.0635$, still over $0.05$, while $135(0.6)^{16} \approx 0.0381$, comfortably under it.

The tolerance sits on the right-hand side of $(0.6)^{\,n} < \text{tolerance}$, and a smaller tolerance is a stricter demand: the powers of $0.6$ have to fall further before the inequality holds. Each additional term multiplies the previous power by $0.6$, which only ever shrinks it further:

$(0.6)^{\,n+1} = 0.6 \cdot (0.6)^{\,n} < (0.6)^{\,n}.$

Since $(0.6)^{\,n}$ is a decreasing function of $n$, reaching a smaller bound takes at least as many steps as reaching a larger one, and in this case strictly more.

That conclusion needed nothing but the shape of the inequality: a stricter bound on a quantity that is monotonically shrinking can only be satisfied later, never sooner, no matter what the specific numbers $135$, $0.6$, or the tolerance itself happen to be. The same reasoning would apply to any convergent geometric series being asked to land within a smaller and smaller distance of its sum.

Divide consecutive terms for the ratio: $\tfrac{15}{40} = \tfrac38$, and $\tfrac{45/8}{15} = \tfrac38$ confirms it. With $a=40$ and $r=\tfrac38$,

$S = \frac{40}{1 - \tfrac38} = \frac{40}{\tfrac58} = 40 \cdot \frac85 = 64.$

Building the partial sums by adding one term at a time:

$S_1 = 40, \quad S_2 = 40+15 = 55, \quad S_3 = 55 + 5.625 = 60.625, \quad S_4 = 60.625 + 2.109375 = 62.734375.$

Each one is bigger than the last, and every one of them is still less than $64$.

Here $r=-\tfrac38$, and

$S = \frac{40}{1 - \left(-\tfrac38\right)} = \frac{40}{\tfrac{11}{8}} = 40 \cdot \frac{8}{11} = \frac{320}{11} \approx 29.09.$

The partial sums now add and subtract in turn:

$S_1 = 40, \quad S_2 = 40 - 15 = 25, \quad S_3 = 25 + 5.625 = 30.625, \quad S_4 = 30.625 - 2.109375 = 28.515625.$

Unlike part A, these hop back and forth: $40$ is above $S$, $25$ is below it, $30.625$ is above it again, $28.515625$ is below it again, each hop smaller than the last.

Everything here comes down to the sign of $r^{\,n}$ in the identity $S - S_n = S\,r^{\,n}$, since $S>0$ is fixed throughout.

Case $0 < r < 1$. A positive number raised to any positive integer power stays positive, so $r^{\,n} > 0$ for every $n$. Multiplying by the positive number $S$ keeps the sign:

$S - S_n = S\,r^{\,n} > 0 \quad\Longrightarrow\quad S_n < S \ \text{ for every } n.$

So no partial sum ever reaches or passes $S$ when the ratio is positive; they climb toward it entirely from below, exactly as part A showed with $S_1 < S_2 < S_3 < S_4 < 64$.

Case $-1 < r < 0$. Now $r$ is negative, and a negative number's power alternates sign with the parity of the exponent: $r^{\,n} > 0$ when $n$ is even, and $r^{\,n} < 0$ when $n$ is odd. So

$S - S_n = S\,r^{\,n} \begin{cases} < 0, & n \text{ odd} \\ > 0, & n \text{ even} \end{cases} \quad\Longrightarrow\quad S_n \begin{cases} > S, & n \text{ odd} \\ < S, & n \text{ even.} \end{cases}$

An odd-numbered partial sum overshoots $S$ and an even-numbered one falls short of it, which is exactly the hopping pattern part B computed: $S_1 = 40 > S$, $S_2 = 25 < S$, $S_3 = 30.625 > S$, $S_4 = 28.515625 < S$. Both cases rest on nothing but the sign of $r^{\,n}$, so the argument holds for every convergent series with $a>0$, not only the two examples above.

Check the hypothesis before trusting the formula. Dividing consecutive terms gives $r = \tfrac{18}{12} = 1.5$, and $|1.5| > 1$, so this series was never eligible for $S = \dfrac{a}{1-r}$ in the first place. The number $-24$ that came out is not a wrong computation, it is a computation that was never asked for: the formula was derived under the assumption $|r|<1$, and outside that assumption there is no reason to expect it to mean anything.

What the series actually does is grow. Every term is $1.5$ times the one before it, and multiplying by a number bigger than $1$ in size, over and over, sends the partial sums

$S_1 = 12, \quad S_2 = 30, \quad S_3 = 57, \quad S_4 = 97.5, \ldots$

past every bound. A sum of ever-growing positive terms cannot possibly equal a negative number, which is itself a warning sign that something upstream had already gone wrong.

The first series has every term equal to $11$, so $r = \tfrac{11}{11} = 1$. Plugging $r=1$ into $S = \dfrac{a}{1-r}$ asks for a division by $1-1=0$, which has no value at all; the formula is not merely inapplicable here, it cannot be evaluated. Directly, for $n$ terms,

$S_n = \underbrace{11 + 11 + \cdots + 11}_{n \text{ terms}} = 11n,$

and $11n$ passes every bound as $n$ grows, so the series diverges by running away.

The second series has ratio $r = \tfrac{12}{16} = \tfrac34$, confirmed by $\tfrac{9}{12} = \tfrac34$, and $\left|\tfrac34\right| < 1$, so this one genuinely converges:

$S = \frac{16}{1 - \tfrac34} = \frac{16}{\tfrac14} = 64.$

Look only at the denominator, $1-r$, since that is the part of the expression that can actually break.

$1 - r = 0 \iff r = 1$

That equation has exactly one solution, so $1-r$ is zero only when $r=1$, and at no other value of $r$: for $r=1.5$, $1-r=-0.5$, a perfectly good nonzero number. So dividing $a$ by $1-r$ is a legitimate arithmetic operation at $r=1.5$, whether or not the result means anything.

That is the whole distinction. At $r=1.5$, the formula runs to completion and hands back a specific number, and the only thing wrong with that number is that it was never entitled to represent a sum, because the derivation that produced the formula assumed $|r|<1$ throughout. At $r=1$, the formula cannot even finish being evaluated, since dividing by zero has no answer, so there is no wrong number to be misled by, only an expression that refuses to be computed. Both situations are dangerous for the same underlying reason, that the condition $|r|<1$ was never checked, but only one of them announces itself by breaking arithmetic outright.