Infinite Geometric Series: Practice

Start from the sum formula and solve for $a$ rather than for $S$. Multiplying both sides of $S = \dfrac{a}{1-r}$ by $1 - r$ gives $a = S(1-r)$.

$a = 30\left(1 - \tfrac25\right) = 30 \cdot \frac35 = 18$

Check it: $\dfrac{18}{1 - 2/5} = \dfrac{18}{3/5} = 30$, as required.

Write out the first terms. At $n = 1$ the exponent is $0$, giving $5$; at $n = 2$ it is $-1$, giving $5 \cdot 2^{-1} = \tfrac52$; at $n = 3$ it is $-2$, giving $\tfrac54$. Each term is half the one before, so $a = 5$ and $r = \tfrac12$.

$S = \frac{5}{1 - \tfrac12} = \frac{5}{\tfrac12} = 10$

A negative exponent means a reciprocal, so $2^{\,1-n}$ shrinks as $n$ grows, which is what makes this converge.

The series is geometric with first term $1$ (which is not zero, so the criterion applies) and common ratio $x$. It converges if and only if the ratio is smaller than $1$ in size.

$|x| < 1 \quad\Longleftrightarrow\quad -1 < x < 1$

Negative values of $x$ are allowed: at $x = -\tfrac12$ the series is $1 - \tfrac12 + \tfrac14 - \cdots$, which converges to $\tfrac23$. That is why $0 < x < 1$ is too small an answer, since it misses every negative $x$ that works. At $x = 0$ the series is $1 + 0 + 0 + \cdots = 1$, which converges too.

The repeating block has two digits, so each block is worth $\tfrac{1}{100}$ of the one before it.

$0.\overline{27} = \frac{27}{100} + \frac{27}{10{,}000} + \cdots = \frac{\tfrac{27}{100}}{1 - \tfrac{1}{100}} = \frac{\tfrac{27}{100}}{\tfrac{99}{100}} = \frac{27}{99}$

Now reduce: $\dfrac{27}{99} = \dfrac{3}{11}$, since both share a factor of $9$. Dividing $3$ by $11$ returns $0.272727\ldots$, confirming the answer.

Two repeating digits means a ratio of $\tfrac{1}{100}$, with first term $\tfrac{45}{100}$.

$0.\overline{45} = \frac{\tfrac{45}{100}}{1 - \tfrac{1}{100}} = \frac{45}{99} = \frac{5}{11}$

Beware of $\tfrac{9}{20}$ and $\tfrac{45}{100}$: both equal $0.45$ exactly, a terminating decimal, which is a different number from $0.454545\ldots$ .

The first rebound height is $\tfrac34$ of the $20$-foot drop, and every later rebound is $\tfrac34$ of the one before it.

$a = 20 \cdot \tfrac34 = 15, \qquad r = \tfrac34$

The rebound heights therefore form the geometric series $15 + \tfrac{45}{4} + \cdots$, and since $\left|\tfrac34\right| < 1$ it converges.

$S = \frac{15}{1 - \tfrac34} = \frac{15}{\tfrac14} = 60$

The original $20$-foot drop is not a rebound, so it is not part of this sum. The trap is to use $a = 20$, which would give $80$.

The exponent is $n$, not $n - 1$, so the first term is not $1$. Substituting $n = 1$ gives $\tfrac23$, and $n = 2$ gives $\tfrac49$, so $a = \tfrac23$ and $r = \tfrac23$.

$S = \frac{\tfrac23}{1 - \tfrac23} = \frac{\tfrac23}{\tfrac13} = 2$

Carelessly using $a = 1$ would give $3$, which is the sum of the series that starts one term earlier, at $n = 0$. Always read the first term off the index.

Put $a = 12$ and $S = 8$ into the sum formula and solve for $r$.

$8 = \frac{12}{1-r} \;\Longrightarrow\; 1 - r = \frac{12}{8} = \frac32 \;\Longrightarrow\; r = 1 - \frac32 = -\frac12$

A negative ratio makes sense here, because the sum $8$ is smaller than the first term $12$, so the later terms must be pulling the total down. Its size is $\tfrac12 < 1$, so the series does converge.

The ratio is the second term over the first, rationalized.

$r = \frac{2\sqrt2}{4} = \frac{\sqrt2}{2} \approx 0.707$

Since $\sqrt2 < 2$, the size of the ratio is less than $1$, so the series converges. Now apply the formula and clear the radical from the denominator by multiplying by the conjugate.

$S = \frac{4}{1 - \tfrac{\sqrt2}{2}} = \frac{8}{2 - \sqrt2} = \frac{8\left(2 + \sqrt2\right)}{\left(2-\sqrt2\right)\left(2+\sqrt2\right)} = \frac{8\left(2+\sqrt2\right)}{2} = 8 + 4\sqrt2$

As a decimal that is about $13.66$, which is comfortably larger than the first few terms added together.

Read the ratio out of each expression. The first has $r = 0.99$, whose size is less than $1$ even though it is close, so it converges (to $\tfrac{3}{0.01} = 300$). The second has $r = -\tfrac12$ and the fourth has $r = \tfrac{1}{100}$, both smaller than $1$ in size.

$\text{Third series: } r = 1, \qquad S_n = \tfrac12 + \tfrac12 + \cdots + \tfrac12 = \frac{n}{2}$

With $r = 1$ every term is the same, so the partial sums are $\tfrac{n}{2}$, and they pass every bound. A large sum is not the same as a divergent one: the first series converges to $300$.

This is a geometric series with $a = 2$ and $r = 1$, and the sum formula is unavailable, because it would divide by $1 - r = 0$. Go back to the partial sums instead.

$S_n = \underbrace{2 + 2 + \cdots + 2}_{n \text{ terms}} = 2n$

Those values $2, 4, 6, 8, \ldots$ pass every bound, so they close in on no number at all and the series diverges. This is the running-away kind of divergence, not the hopping kind, which is what $r = -1$ would produce.

The first term is $4$, which is not zero, so the criterion applies: the series converges exactly when the ratio is smaller than $1$ in size. Here the ratio is $\tfrac{x}{3}$.

$\left|\frac{x}{3}\right| < 1 \;\Longleftrightarrow\; |x| < 3 \;\Longleftrightarrow\; -3 < x < 3$

The answer $x < 3$ is wrong because it lets in numbers like $x = -10$, whose ratio $-\tfrac{10}{3}$ has size greater than $1$, so that series diverges.