Infinite Geometric Series: Practice

A geometric series with a nonzero first term converges precisely when the size of its ratio is less than $1$, so find $r$ in each case.

The first has $r = 2$, the second has $r = 1$, and the fourth has $r = -1$, so all three fail the test. The fourth is the interesting failure: its partial sums are $1, 0, 1, 0, \ldots$, which stay bounded and still never settle.

$\text{Third series: } r = \frac{3}{6} = \frac12, \qquad \left|\tfrac12\right| < 1$

Only the third has a ratio smaller than $1$ in size, so only it converges.

The first term is $a = 8$ and the ratio is $r = \tfrac48 = \tfrac12$. Since $\left|\tfrac12\right| < 1$, the series converges and the sum formula applies.

$S = \frac{a}{1-r} = \frac{8}{1 - \tfrac12} = \frac{8}{\tfrac12} = 16$

The partial sums $8, 12, 14, 15, 15.5, \ldots$ are climbing toward $16$, which is a quick sanity check.

The ratio satisfies $\left|\tfrac13\right| < 1$, so the series converges and $S = \dfrac{a}{1-r}$ with $a = 6$ and $r = \tfrac13$.

$S = \frac{6}{1 - \tfrac13} = \frac{6}{\tfrac23} = 6 \cdot \frac32 = 9$

The common wrong turn is to divide by $r$ instead of by $1 - r$, which would give $18$.

With a nonzero first term, the series converges if and only if $|r| < 1$, so compare the size of each ratio with $1$.

The sizes are $0.9$, $0.5$, $1.2$, and $0.25$. Three of them are below $1$, and $0.9$ is below $1$ even though it is close to it, so those three series converge.

$|1.2| = 1.2 > 1$

Only $r = 1.2$ fails the test. Its powers $(1.2)^n$ grow past every bound, so the partial sums run away and the series diverges.

Here $a = 1$ and $r = \tfrac13$, since each term is one third of the one before it. The ratio is smaller than $1$ in size, so the series converges.

$S = \frac{1}{1 - \tfrac13} = \frac{1}{\tfrac23} = \frac32$

The partial sums $1, 1.\overline{3}, 1.\overline{4}, \ldots$ close in on $1.5$ from below.

This is the cake from the lesson: $a = \tfrac12$ and $r = \tfrac12$.

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

Every partial sum ($\tfrac12$, $\tfrac34$, $\tfrac78$, $\ldots$) is less than $1$, and the sum is exactly $1$, because the sum is the number the partial sums close in on, not one of them.

Since $\left|\tfrac15\right| < 1$, the series converges. Apply the formula with $a = 10$ and $r = \tfrac15$.

$S = \frac{10}{1 - \tfrac15} = \frac{10}{\tfrac45} = 10 \cdot \frac54 = \frac{25}{2}$

So the sum is $12.5$. Dividing by $r$ rather than $1 - r$ would give $50$, which is the trap here.

The sum of a series is defined to be the number its partial sums close in on, if there is one. Each partial sum here adds one more $9$, so the gap up to $1$ is $0.1$, then $0.01$, then $0.001$, shrinking past any tolerance you name.

$1 - S_n = \left(\tfrac{1}{10}\right)^{n} \quad\text{closes in on}\quad 0$

So the partial sums close in on $1$, and the sum is exactly $1$. No partial sum equals $1$, and that is fine: the sum is not one of the partial sums.

Divide the second term by the first.

$r = \frac{4/25}{4/5} = \frac{4}{25} \cdot \frac{5}{4} = \frac15$

The denominators $5, 25, 125$ multiply by $5$ each step, so the terms are divided by $5$ each step. The ratio is $\tfrac15$, not $5$.

The partial sums split as $S_n = \dfrac{a}{1-r} - \dfrac{a}{1-r}\,r^{\,n}$, so the formula is the number they close in on only when the powers $r^{\,n}$ die away.

$r^{\,n} \text{ closes in on } 0 \quad\text{exactly for}\quad |r| < 1$

The sign of $a$ and the sign of $r$ are irrelevant; the series $1 - \tfrac12 + \tfrac14 - \cdots$ has a negative ratio and converges perfectly well, so $r > 0$ and $a > 0$ are not required. What matters is that the ratio is smaller than $1$ in size. Applied outside that range the formula returns nonsense, such as $1 + 2 + 4 + \cdots = \tfrac{1}{1-2} = -1$.

The stem rules out a first term of $0$ for a reason. The all-zeros series converges to $0$ for every ratio, and the formula returns $\tfrac{0}{1-r} = 0$ there too, so without the nonzero hypothesis the words "only when" would be false.

The series is written in the standard shape $a + ar + ar^2 + \cdots$, so read off $a = 20$ and $r = 0.1$. Since $|0.1| < 1$ it converges.

$S = \frac{20}{1 - 0.1} = \frac{20}{0.9} = \frac{200}{9}$

That is $22.\overline{2}$, which is close to $22$ but not equal to it. The partial sums $20, 22, 22.2, 22.22, \ldots$ show exactly where the repeating $2$s come from.

Write out the first few terms by substituting $n = 1, 2, 3$: they are $3$, $3\left(\tfrac14\right) = \tfrac34$, and $3\left(\tfrac14\right)^2 = \tfrac{3}{16}$. So $a = 3$ and $r = \tfrac14$.

$S = \frac{3}{1 - \tfrac14} = \frac{3}{\tfrac34} = 4$

The exponent $n-1$ is what makes the first term plain $a$: at $n = 1$ the power is $\left(\tfrac14\right)^0 = 1$.