Infinite Geometric Series: Practice

Treat $x$ as the first term. The ratio is $\tfrac12$, whatever $x$ happens to be, so the series converges and the sum formula applies.

$S = \frac{x}{1 - \tfrac12} = \frac{x}{\tfrac12} = 2x$

An infinite series with ratio $\tfrac12$ therefore totals exactly twice its first term. Setting that equal to $10$,

$2x = 10 \;\Longrightarrow\; x = 5.$

The error after $n$ terms is exactly $S\,r^{\,n} = 6(0.5)^n$, so the demand is an inequality you can solve.

$6(0.5)^n < 0.01 \;\Longleftrightarrow\; (0.5)^n < \frac{1}{600}$

Take logarithms. Since $\log(0.5) \approx -0.30103$ is negative, dividing by it reverses the inequality.

$n > \frac{\log(1/600)}{\log(0.5)} = \frac{-2.778}{-0.301} \approx 9.23$

So $n = 10$ is the first term count that works. Checking the boundary confirms it: $6(0.5)^9 = \tfrac{6}{512} \approx 0.0117$, still too big, while $6(0.5)^{10} = \tfrac{6}{1024} \approx 0.0059$, inside the tolerance.

Every first term here is nonzero, so the test is purely the size of the ratio. The first has $\left|\tfrac{\sqrt2}{2}\right| \approx 0.707$, the second has $0.999$, and the fourth has $0.99$. All three are strictly below $1$, so all three converge, however slowly.

$\text{Third series: } r = -1, \qquad |r| = 1$

A ratio of exactly $-1$ fails the test. Its terms are $3, -3, 3, -3, \ldots$ and its partial sums are $3, 0, 3, 0, \ldots$, which never settle. Note that being close to $1$ is not the same as reaching it: $0.999$ still converges, to $\tfrac{1}{0.001} = 1000$.

Each new square has side $\tfrac{s}{\sqrt2}$, so the perimeters shrink by the factor $\tfrac{1}{\sqrt2} = \tfrac{\sqrt2}{2}$, not by $\tfrac12$. The first perimeter is $4 \cdot 8 = 32$.

$a = 32, \qquad r = \frac{\sqrt2}{2} \approx 0.707 < 1$

Apply the formula and rationalize the denominator with the conjugate.

$S = \frac{32}{1 - \tfrac{\sqrt2}{2}} = \frac{64}{2 - \sqrt2} = \frac{64\left(2+\sqrt2\right)}{2} = 64 + 32\sqrt2$

That is roughly $109.3$. Areas shrink by $\tfrac12$ while perimeters shrink only by $\tfrac{\sqrt2}{2}$, so the perimeters pile up more slowly and the total is comparatively larger.

The first term is $(2x-1)^0 = 1$, which is never zero, so the criterion applies with common ratio $r = 2x - 1$. The series converges exactly when that ratio is smaller than $1$ in size.

$|2x - 1| < 1 \;\Longleftrightarrow\; -1 < 2x - 1 < 1$

Add $1$ throughout, then divide by $2$.

$0 < 2x < 2 \;\Longleftrightarrow\; 0 < x < 1$

Test an endpoint to see why it is excluded: at $x = 1$ the ratio is $1$ and the series is $1 + 1 + 1 + \cdots$, which runs away.

The repeating block has three digits, so each block is worth $\tfrac{1}{1000}$ of the one before it.

$0.\overline{123} = \frac{123}{1000} + \frac{123}{10^6} + \cdots = \frac{\tfrac{123}{1000}}{1 - \tfrac{1}{1000}} = \frac{123}{999}$

Now reduce. Both $123 = 3 \cdot 41$ and $999 = 3 \cdot 333$ carry a factor of $3$.

$\frac{123}{999} = \frac{41}{333}$

Since $41$ is prime and does not divide $333 = 3 \cdot 3 \cdot 37$, this is lowest terms.

First recover the ratio of the original series from $\dfrac{2}{1-r} = 6$.

$1 - r = \frac26 = \frac13 \;\Longrightarrow\; r = \frac23$

The original terms are $2, 2r, 2r^2, \ldots$, so their squares are $4, 4r^2, 4r^4, \ldots$, which is again geometric, with first term $4$ and ratio $r^2 = \tfrac49$. That ratio is still smaller than $1$, so the squared series converges too.

$S = \frac{4}{1 - \tfrac49} = \frac{4}{\tfrac59} = \frac{36}{5}$

Squaring the terms does not square the sum: $6^2 = 36$ is not the answer.

The steps are $50$, then $25$, then $12.5$, and so on, each half the one before, so they form a geometric series with $a = 50$ and $r = \tfrac12$.

$S = \frac{50}{1 - \tfrac12} = \frac{50}{\tfrac12} = 100$

The steps total exactly the width of the room. There are infinitely many of them, and their sum is finite, and the two facts sit together without any contradiction: the sum of a series is the number its partial sums close in on, not the outcome of performing infinitely many additions one after another.

With $r = -1$ the terms are $a, -a, a, -a, \ldots$, and each new term undoes the one before it.

$S_1 = a, \quad S_2 = 0, \quad S_3 = a, \quad S_4 = 0, \quad \ldots$

Because $a \neq 0$, the values $a$ and $0$ are different, so the partial sums hop between two distinct numbers forever. They never grow (every one of them is $a$ or $0$, so they stay bounded), and they never settle on a single value, so the series diverges. This is the failure mode that shows divergence and blowing up are not the same thing. The value $\tfrac{a}{2}$ is what the sum formula would return if it were legal to use here, and it is the sum of nothing at all.

Substituting $n = 1$ gives the first term $\dfrac{3^0}{4^2} = \dfrac{1}{16}$. For the ratio, notice that each step multiplies the numerator by $3$ and the denominator by $4$.

$r = \frac{3}{4}$

It is worth seeing this structurally: $\dfrac{3^{\,n-1}}{4^{\,n+1}} = \dfrac{1}{16}\left(\dfrac34\right)^{\,n-1}$, since $4^{\,n+1} = 16 \cdot 4^{\,n-1}$. So the series is an ordinary geometric one with $a = \tfrac{1}{16}$.

$S = \frac{\tfrac{1}{16}}{1 - \tfrac34} = \frac{\tfrac{1}{16}}{\tfrac14} = \frac14$

The error after $n$ terms is exactly $S\,r^{\,n} = 50(0.8)^n$, so solve the inequality that says the error is small enough.

$50(0.8)^n < 1 \;\Longleftrightarrow\; (0.8)^n < 0.02$

Take logarithms, then divide by $\log(0.8) \approx -0.0969$, which flips the inequality because it is negative.

$n > \frac{\log(0.02)}{\log(0.8)} = \frac{-1.699}{-0.0969} \approx 17.53$

So the fewest that works is $n = 18$, and every larger $n$ works too, which is why the question asks for the fewest. The boundary check is quick: $50(0.8)^{17} \approx 1.13$, still outside the tolerance, while $50(0.8)^{18} \approx 0.90$, inside it.

The quickest route uses the error formula: what is left after $n$ terms is exactly $S\,r^{\,n}$, and here $n = 3$.

$S - S_3 = 6\left(\tfrac13\right)^3 = \frac{6}{27} = \frac29$

The slower route confirms it. The first three terms are $4$, $\tfrac43$, and $\tfrac49$, which add to $\tfrac{36 + 12 + 4}{9} = \tfrac{52}{9}$, so the leftover is $6 - \tfrac{52}{9} = \tfrac{54 - 52}{9} = \tfrac29$.

You can also sum the tail directly as a new geometric series: its first term is $4\left(\tfrac13\right)^3 = \tfrac{4}{27}$ and its ratio is still $\tfrac13$, giving $\dfrac{4/27}{2/3} = \tfrac29$ again. Note that $\tfrac{4}{27}$ by itself is only the first leftover term, not the whole tail.