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Infinite Geometric Series: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A fixed opening term

Find the exact value of $2+\sum_{k=1}^{\infty}\frac3{2^k}$ .
1. Hint 1
  Separate the fixed opening term from the geometric tail.
2. Hint 2
  The tail starts with $\frac32$ and has ratio $\frac12$ .
Answer

$5$ .

Full solution

The tail converges because $|r|=\frac12$ , which is less than $1$ .
Its sum is
$\frac{3/2}{1-1/2}=3$
Adding the fixed opening $2$ gives $5$ .
The opening term is not the first term used in the tail's geometric formula.

Answer

$5$ .

Key idea

A finite opening part can be added to a convergent geometric tail after the tail is identified correctly.
Problem 1: Mark complete
Problem 2 A squared ratio

For which real $x$ does $1+\frac{x^2}{4}+(\frac{x^2}{4})^2+\cdots$ converge?
1. Hint 1
  The first term is nonzero, so convergence requires the ratio's absolute value to be less than one.
2. Hint 2
  The ratio $x^2/4$ is already zero or positive.
Answer

$-2<x<2$ .

Full solution

The criterion becomes $x^2/4<1$ , or
$x^2<4$
Thus $-2<x<2$ .
At either endpoint the ratio is one, and beyond the endpoints it exceeds one.
At $x=0$ the series is $1+0+0+\cdots$ , which also converges.

Answer

$-2<x<2$ .

Key idea

A convergence condition can become a quadratic inequality when the ratio depends on a squared parameter.
Problem 2: Mark complete
Problem 3 Two repeating decimals

Write $u=0.\overline{62}$ and $v=0.6\overline{2}$ as fractions in lowest terms. Which is larger, and by exactly how much?
1. Hint 1
  Locate the repeating block in each number.
2. Hint 2
  The block shifts two places each time in u and one place each time in v.
3. Hint 3
  Separate any nonrepeating opening before summing the geometric tail.
Answer

$u=\frac{62}{99}$ and $v=\frac{28}{45}$ ; $u$ is larger by $\frac2{495}$ .

Full solution

For $u$ , the blocks $0.62$ , $0.0062$ , and so on form a geometric series with first term $\frac{62}{100}$ and ratio $\frac1{100}$ , so $u=\frac{62/100}{99/100}$ , which is $\frac{62}{99}$ .

For $v$ , the opening is $\frac6{10}$ and the tail $0.02+0.002+\cdots$ has first term $\frac2{100}$ and ratio $\frac1{10}$ , so the tail is $\frac2{90}$ and $v=\frac{54}{90}+\frac2{90}$ , which is $\frac{28}{45}$ .

Over the common denominator $495$ , $u=\frac{310}{495}$ and $v=\frac{308}{495}$ , so $u$ is larger by $\frac2{495}$ .

Answer

$u=\frac{62}{99}$ and $v=\frac{28}{45}$ ; $u$ is larger by $\frac2{495}$ .

Key idea

Repeating decimals that look alike can differ once each is written as a geometric series and summed exactly.
Problem 3: Mark complete
Problem 4 A sequence of corrections

A measurement starts at $10$ units. Corrections of $3,-1,\frac13,-\frac19,\ldots$ units are then added in order, each correction being $-\frac13$ times the previous one. What value do the corrected readings approach?
1. Hint 1
  The corrected reading is the starting measurement plus a partial sum of corrections.
2. Hint 2
  The correction series has ratio of size below one, so it has a finite sum.
Answer

$\frac{49}{4}$ units, or $12.25$ units.

Full solution

The correction series has $a=3$ and $r=-\frac13$ , so its sum is
$S=\frac3{1+1/3}$
or $S=\frac94$ units.

Adding the initial $10$ units gives $10+\frac94=\frac{49}{4}$ units.
The first readings $13,12,12\frac13$ lie on alternating sides of $12.25$ , with shrinking corrections.

Answer

$\frac{49}{4}$ units, or $12.25$ units.

Key idea

Adding a convergent correction series shifts the limiting reading by the series' sum.
Problem 4: Mark complete
Problem 5 Keeping alternate terms

A geometric series has terms $a_n=14(-\frac25)^{n-1}$ for positive integers $n$ . Form a new series using just the terms with even indices, $a_2+a_4+a_6+\cdots$ . Find its first term, ratio, and sum.
1. Hint 1
  Successive retained terms are two original ratio steps apart.
2. Hint 2
  The retained series starts at the second original term, not the first.
Answer

First term $-\frac{28}5$ ; ratio $\frac4{25}$ ; sum $-\frac{20}3$ .

Full solution

The first retained term is $a_2=14(-\frac25)=-\frac{28}5$ , and each retained term is two ratio steps after the last, so the ratio is
$(-\frac25)^2=\frac4{25}.$

The ratio has size below one, so
$S=\frac{-28/5}{1-4/25}$
giving
$S=-\frac{20}3.$
The odd-indexed terms sum to $\frac{14}{21/25}=\frac{50}3$ , and $\frac{50}3-\frac{20}3=10$ checks against the original total
$\frac{14}{1+2/5}=10.$

Answer

First term $-\frac{28}5$ ; ratio $\frac4{25}$ ; sum $-\frac{20}3$ .

Key idea

Keeping the even-indexed terms of a geometric series starts the new series at the second term and squares the common ratio.
Problem 5: Mark complete
Problem 6 A tail and its opening

A convergent geometric series has first term $a\ne0$ and ratio $r=\frac34$ . Its sum after the first two terms are removed is $27$ . Find $a$ and the sum of the full series.
1. Hint 1
  The remaining series starts at $ar^2$ and keeps the same ratio.
2. Hint 2
  Express its sum in terms of $a$ , then restore the original starting point.
Answer

$a=12$ ; the full sum is $48$ .

Full solution

The tail begins with $a(\frac34)^2$ , so its sum is
$27=\frac{a(9/16)}{1/4}$
This simplifies to $27=9a/4$ , giving $a=12$ .

The full sum is
$12/(1-\frac34)=48.$
Its first two terms are $12$ and $9$ , whose sum $21$ leaves the given tail sum $27$ .

Answer

$a=12$ ; the full sum is $48$ .

Key idea

Removing initial terms shifts the first term of a geometric tail without changing its ratio.
Problem 6: Mark complete
Problem 7 Two estimates to compare

For the series $6+1.5+0.375+\cdots$ , one estimate of its infinite sum uses the first three terms. Another estimate is $7.9$ . Which estimate is closer to the sum, and by exactly how much is its absolute error smaller?
1. Hint 1
  Identify the common ratio and check that an infinite sum exists.
2. Hint 2
  Find the exact sum and compare its distances from the two estimates using fractions.
Answer

The estimate $7.9$ is closer; its absolute error is smaller by $\frac1{40}$ .

Full solution

Each term is one quarter of the preceding term, so the series converges.
Its exact sum is
$S=\frac{6}{1-1/4}$
Thus $S=8$ .

The first three terms total $6+\frac32+\frac38=\frac{63}{8}$ , whose error is
$8-\frac{63}{8}=\frac18.$
The other estimate is $\frac{79}{10}$ , whose error is $\frac1{10}$ .

The second error is smaller, and the difference is
$\frac18-\frac1{10}=\frac1{40}$
Both estimates lie below the true sum, so these differences are their absolute errors.

Answer

The estimate $7.9$ is closer; its absolute error is smaller by $\frac1{40}$ .

Key idea

Comparing approximations requires their distances from the exact sum, regardless of how each approximation was produced.
Problem 7: Mark complete
Problem 8 Two families of partial sums

Two series have partial sums $P_n=cn$ and $Q_n=c+(-1)^n$ for positive integers $n$ , where $c$ is a real constant. Find every value of $c$ for which each series converges. For each divergent case, state whether its partial sums are bounded.
1. Hint 1
  A series converges exactly when its partial sums settle on one number.
2. Hint 2
  Consider c = 0 separately for the first series; for the second, compare odd and even n.
Answer

$P$ : converges (to $0$ ) only for $c=0$ , and is unbounded for $c\ne0$ . $Q$ : diverges for every $c$ , with bounded partial sums.

Full solution

If $c=0$ , every $P_n$ is $0$ , so the first series converges to $0$ .
If $c\ne0$ , then $|P_n|=|c|n$ passes every bound, so it diverges with unbounded partial sums.

For every $c$ , $Q_n$ is $c-1$ at odd $n$ and $c+1$ at even $n$ .
These two values differ by $2$ , so the partial sums never settle and the series diverges, yet they stay between $c-1$ and $c+1$ , so they are bounded.

Answer

$P$ : converges (to $0$ ) only for $c=0$ , and is unbounded for $c\ne0$ . $Q$ : diverges for every $c$ , with bounded partial sums.

Key idea

Divergence means the partial sums fail to settle on one value, which can happen with bounded or with unbounded partial sums.
Problem 8: Mark complete
Problem 9 A gap that never closes

For real $a\ne0$ and $0<|r|<1$ , let $S$ be the sum of $a+ar+ar^2+\cdots$ and let $S_n$ be the sum of its first $n$ terms. Show that $S-S_n=Sr^n$ , and decide whether $S_n=S$ for some positive integer $n$ .
1. Hint 1
  Write both S and the partial sum with the same denominator.
2. Hint 2
  Decide when a product of two real numbers can be zero.
Answer

$S-S_n=Sr^n$ ; no, $S_n\ne S$ for every positive integer $n$ .

Full solution

Here $S=\frac a{1-r}$ and $S_n=\frac{a(1-r^n)}{1-r}$ , so
$S-S_n=\frac{ar^n}{1-r}$
which is $Sr^n$ .

Since $a\ne0$ , $S\ne0$ , and since $r\ne0$ , $r^n\ne0$ .
So the gap $Sr^n$ is never zero and no partial sum equals $S$ .
The gap still shrinks toward $0$ because $|r|<1$ , which is why $S$ is the sum.

Answer

$S-S_n=Sr^n$ ; no, $S_n\ne S$ for every positive integer $n$ .

Key idea

With a nonzero first term and ratio, no partial sum of a convergent geometric series equals its sum, although the gap shrinks toward zero.
Problem 9: Mark complete
Problem 10 A sign-changing claim

For real $a\ne0$ and $-1<r<0$ , a student says $a+ar+ar^2+\cdots$ converges even though its terms change sign. Is that true? Explain why the changing signs do not prevent the partial sums from settling.
1. Hint 1
  Convergence is controlled by the size of the ratio.
2. Hint 2
  The exact error contains $r^n$ , whose magnitude is $|r|^n$ .
Answer

Yes; the sum is $\frac{a}{1-r}$ .

Full solution

The conditions imply $|r|<1$ , so $|r|^n$ approaches zero.
With $S=a/(1-r)$ , the partial sums satisfy
$|S-S_n|=|S||r|^n$
The error magnitude tends to zero, while the alternating sign merely places successive partial sums on opposite sides of $S$ .
Thus they settle on the same value.

Answer

Yes; the sum is $\frac{a}{1-r}$ .

Key idea

Alternating signs are compatible with geometric convergence when the ratio's magnitude is below one.
Problem 10: Mark complete

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