MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: geometric series, telescoping series and the divergence test (MATH 141)

Q: How do I find the sum of a geometric series that starts at n = 1?

Compute the first term by substituting the starting index, then divide it by one minus the ratio, provided the absolute value of the ratio is less than one. For the series of two thirds to the power n starting at n equals 1, the first term is two thirds and the sum is 2, not 3. One over one minus r only works when the first term is 1.

Q: If the terms of a series go to zero, does the series converge?

Not necessarily. Terms tending to zero is necessary for convergence but not sufficient. The harmonic series, the sum of one over n, has terms tending to zero and still diverges, because its partial sums grow without bound. The divergence test can only prove divergence; when the limit of the terms is zero, it tells you nothing.

Q: How do I know how many terms survive in a telescoping series?

Write out the partial sum with its first three brackets and its last two. If each negative piece cancels with the positive piece of the next bracket, one term survives at each end; if it cancels two brackets later, two terms survive at each end. Never assume the cancellation: the written brackets are what the marker wants to see.

Q: Can I split a series into two sums when finding its value?

Only when both resulting series converge. Splitting the sum of one over n minus one over n plus 1 into two harmonic series gives infinity minus infinity, which has no value. Work on the finite partial sum instead, where every rearrangement is allowed, and take the limit at the very end.

Q: Why is 0.999 repeating equal to 1?

Because the decimal is the geometric series nine tenths plus nine hundredths plus nine thousandths and so on, with first term nine tenths and ratio one tenth. Its sum is nine tenths divided by nine tenths, which is exactly 1. Every partial sum is less than 1, but the decimal means the limit of those partial sums, and that limit is 1.

This sheet is not a summary of section 11.2 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on the first series chapter of MATH 141 at McGill University, and which precise gesture avoids each loss.

Everything below is done by hand, as on the exam, and every sum quoted has been checked against its partial sums. Two sequences live in every question of the chapter, the terms $a_n$ and the partial sums $s_n$ ; nearly every trap on this page is a confusion between them.

The thread of the chapter

A series is the limit of its PARTIAL SUMS $s_n$ , never of its terms: write $s_n$ (geometric closed form, telescoping cancellation, grouping) before concluding. Terms that tend to $0$ are necessary for convergence and never sufficient.

This chapter is part of MATH 141, Calculus 2 (McGill)

The essentials

Two sequences in every question: the terms and the partial sums

• The terms: $a_1, a_2, a_3, \dots$ The partial sums: $s_n = a_1 + \dots + a_n$ . The series $\sum a_n$ CONVERGES when $(s_n)$ has a finite limit $s$ , its sum.
• From one to the other: $s_n = s_{n-1} + a_n$ , so $a_1 = s_1$ and $a_n = s_n - s_{n-1}$ for $n \ge 2$ .
• If $\sum a_n$ converges then $a_n \to 0$ (because $a_n = s_n - s_{n-1} \to s - s$ ). The converse is FALSE: $\frac{1}{n} \to 0$ and $\sum \frac{1}{n}$ diverges.
• Hence the divergence test: $\lim a_n \ne 0$ , or no limit, proves divergence. $\lim a_n = 0$ proves nothing.
• Positive terms: $(s_n)$ increases, so the sum is larger than every partial sum, in particular larger than $a_1$ .

The terms

\frac{1}{2^n}

and

\frac{1}{n}

both tend to

0

, yet the partial sums of the first level off at

1

while those of the second climb forever.

Before writing a limit, write which sequence it is the limit of. $\lim a_n$ is a test, $\lim s_n$ is the answer.

The two series you can actually sum

• Geometric: each term is the previous one times the same $r = \frac{a_{n+1}}{a_n}$ . If $|r| < 1$ , the sum is $\frac{\text{first term}}{1 - r}$ ; if $|r| \ge 1$ (first term not $0$ ), it diverges.
• The first term is the term at the STARTING index, $n = 0$ , $1$ or $2$ : compute it by substitution, never read it off the formula.
• Telescoping: $\sum_{n=1}^{\infty} (b_{n+1} - b_n)$ has $s_N = b_{N+1} - b_1$ . It converges exactly when $b_n$ has a finite limit $L$ , to $L - b_1$ .
• The difference is found by partial fractions ( $\frac{1}{n(n+2)} = \frac{1}{2}(\frac{1}{n} - \frac{1}{n+2})$ ), by logarithms ( $\ln\frac{n}{n+1} = \ln n - \ln(n+1)$ ) or by rewriting a numerator ( $n = (n+1) - 1$ ).
• Linearity holds for CONVERGENT series only: $\sum(a_n + b_n) = \sum a_n + \sum b_n$ when both converge.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Which verdict, from which form of the series

Read a line as: when the series has the form of the first column and satisfies the condition, the verdict is the last column. Each line carries a computed example; the red lines are the conclusions students invent.

Form of the series	Condition	Verdict
geometric, ratio $r$	$\|r\| < 1$	converges to $\frac{\text{first term}}{1 - r}$
Example: $\sum_{n=2}^{\infty} \frac{(-3)^{n+1}}{4^n}$ : first term $-\frac{27}{16}$ , $r = -\frac{3}{4}$ , sum $-\frac{27}{28}$ .
geometric, ratio $r$	$\|r\| \ge 1$	diverges
Example: $\sum \frac{\pi^n}{3^{n+1}}$ : $r = \frac{\pi}{3} > 1$ since $\pi > 3$ .
telescoping $\sum (b_{n+1} - b_n)$	$b_n \to L$ finite	converges to $L - b_1$
Example: $\sum (\arctan(n+1) - \arctan n) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$ .
telescoping $\sum (b_{n+1} - b_n)$	$b_n$ has no finite limit	diverges
Example: $\sum (\sqrt{n+1} - \sqrt{n})$ : $s_N = \sqrt{N+1} - 1 \to \infty$ .
any $\sum a_n$	$\lim a_n \ne 0$ or no limit	diverges
Example: $\sum n\sin\frac{1}{n}$ : $a_n \to 1$ . $\sum \cos(n\pi)$ : $a_n = (-1)^n$ has no limit.
any $\sum a_n$	$\lim a_n = 0$	no conclusion no conclusion
Example: $\sum \frac{1}{n^2 + n}$ converges to $1$ and $\sum \frac{1}{n}$ diverges: both have $a_n \to 0$ . Same form, other result: $\sum \ln\frac{n}{n+1}$ , terms $\to 0$ , $s_N = -\ln(N+1) \to -\infty$ ; $\sum \frac{2}{3^n}$ , terms $\to 0$ , $s_n \to 1$ . What to do: Compute $s_n$ (telescoping, bounding, grouping), or use the tests of the next chapters.
$\sum (a_n + b_n)$	both diverge	no rule no rule
Example: $a_n = 1$ , $b_n = -1$ : $\sum 0 = 0$ converges. $a_n = b_n = 1$ : $\sum 2$ diverges. What to do: Combine the terms first, then study the single series $\sum (a_n + b_n)$ on its own partial sums.

No line of this table concludes convergence without a closed form for $s_n$ : in this chapter, a convergence verdict always comes with its sum.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Using 1 over 1 minus r on a series that starts at n = 1

the whole numerical answer, typically 2 to 3 marks

What not to write

“ $\sum_{n=1}^{\infty} \left(\frac{2}{3}\right)^n = \frac{1}{1 - 2/3} = 3$ .”

What to write

“First term ( $n = 1$ ): $\frac{2}{3}$ ; ratio $\frac{2}{3}$ ; sum $\frac{2/3}{1 - 2/3} = 2$ .”

The partial sums of

\sum_{n \ge 1} \left(\frac{2}{3}\right)^n

climb to

2

; the dashed line at

3

is what

\frac{1}{1 - r}

gives, a level the points never approach.

Why: $\frac{1}{1 - r}$ is the sum of $1 + r + r^2 + \dots$ , whose first term is $1$ . The formula to memorize is first term over $1 - r$ , and the first term is obtained by substituting the starting index.

2. Reading the ratio off the letter under the exponent

the whole question: the verdict itself is reversed

What not to write

“ $\sum_{n=0}^{\infty} \frac{2^{2n+1}}{3^n}$ : ratio $\frac{2}{3}$ , so it converges to $\frac{2}{1 - 2/3} = 6$ .”

What to write

“ $2^{2n+1} = 2 \cdot 4^n$ , so the term is $2\left(\frac{4}{3}\right)^n$ : $r = \frac{4}{3} \ge 1$ , the series diverges.”

Why: The ratio is $\frac{a_{n+1}}{a_n}$ , and an exponent $2n$ squares the base. Rewrite every term as (constant) times (number) $^n$ before naming $r$ . The same care applies to $n + 1$ or $n - 1$ in an exponent, which change the first term, not the ratio.

3. Concluding convergence from terms that tend to 0

all the marks of the question: a wrong test counts as no argument

What not to write

“ $\lim \frac{1}{n} = 0$ , so by the divergence test $\sum \frac{1}{n}$ converges.”

What to write

“ $\lim \frac{1}{n} = 0$ : the divergence test is inconclusive. Grouping gives $s_{2^k} \ge 1 + \frac{k}{2}$ , so the harmonic series diverges.”

Why: The divergence test has only one verdict, diverges. $a_n \to 0$ is necessary for convergence and far from sufficient: $\sum \frac{1}{n}$ , $\sum \ln\frac{n}{n+1}$ and $\sum (\sqrt{n+1} - \sqrt{n})$ all have vanishing terms and all diverge.

4. Taking the limit of the terms for the sum

2 marks, and the credibility of everything around it

What not to write

“The partial sums are $s_n = \frac{4n}{n+1}$ , so $a_n = \frac{4}{n(n+1)}$ and the sum is $\lim a_n = 0$ .”

What to write

“The sum is $\lim s_n = \lim \frac{4n}{n+1} = 4$ .”

Why: $\lim a_n$ is $0$ for EVERY convergent series, so it cannot be the sum of any of them. A series of positive terms has a sum at least $a_1$ , here $2$ : the answer $0$ fails that check at once.

5. Splitting a telescoping series into two divergent series

the whole question, since the verdict is wrong

What not to write

“ $\sum \left(\frac{1}{n} - \frac{1}{n+1}\right) = \sum \frac{1}{n} - \sum \frac{1}{n+1} = \infty - \infty$ , so it diverges.”

What to write

“ $s_N = \left(1 - \frac{1}{2}\right) + \dots + \left(\frac{1}{N} - \frac{1}{N+1}\right) = 1 - \frac{1}{N+1} \to 1$ .”

Why: $\sum(a_n - b_n) = \sum a_n - \sum b_n$ requires both series to converge. Work on the FINITE sum $s_N$ , where any rearrangement is legal, and take the limit only at the end.

6. Keeping one surviving term when the gap is two

1 to 2 marks

What not to write

“ $\frac{1}{n(n+2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right)$ , everything cancels except the first term, so the sum is $\frac{1}{2}$ .”

What to write

“ $s_N = \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{N+1} - \frac{1}{N+2}\right) \to \frac{3}{4}$ .”

Why: When the difference is between $b_n$ and $b_{n+2}$ , each term cancels with one two places further on, and two terms survive at each end. Writing the first three brackets and the last two is what shows it; the cancellation is never assumed.

7. Inserting brackets into a divergent series

1 to 2 marks

What not to write

“ $1 - 1 + 1 - 1 + \dots = (1 - 1) + (1 - 1) + \dots = 0$ .”

What to write

“The terms $(-1)^n$ do not tend to $0$ , so the series diverges; its partial sums $1, 0, 1, 0, \dots$ have no limit.”

Why: Brackets replace $(s_n)$ by one of its subsequences. That is harmless when $s_n$ converges and meaningless otherwise: grouping as $1 + (-1 + 1) + \dots$ gives $1$ instead of $0$ .

8. Using the geometric formula when the ratio is at least 1

the whole question

What not to write

“ $\sum_{n=0}^{\infty} 2\left(\frac{4}{3}\right)^n = \frac{2}{1 - 4/3} = -6$ .”

What to write

“ $|r| = \frac{4}{3} \ge 1$ : the series diverges; its terms do not even tend to $0$ .”

Why: $\frac{a}{1 - r}$ comes from $s_n = a\frac{1 - r^n}{1 - r}$ and the fact that $r^n \to 0$ , which requires $|r| < 1$ . Check $|r| < 1$ in writing before using the formula; a negative sum of positive terms is the alarm.

Which method to choose

What to do with a series, by the FORM of its general term

Look at the term a_n before choosing anything, and ask the first question first

If $a_n$ does not tend to $0$ (a rational function of equal degrees, $n\sin\frac{1}{n}$ , $(-1)^n$ ) → diverges by the divergence test; stop there

Example: $\sum \frac{2n^2 - 1}{5n^2 + n}$ : $a_n \to \frac{2}{5}$
If $a_n$ is (constant) times (number) $^n$ , possibly after rewriting $2^{2n}$ or $5^{n-1}$ → geometric: first term by substitution, ratio by quotient, then |r| < 1 or not

Example: $\frac{4^n}{5^{n-1}} = 4\left(\frac{4}{5}\right)^{n-1}$ : sum $20$
If $a_n$ is a rational function with a denominator of shifted factors, $n(n+1)$ , $n(n+2)$ , $n^2 - 1$ → partial fractions, then write s_N and count the surviving terms

Example: $\frac{1}{n(n+2)}$ : sum $\frac{3}{4}$
If $a_n$ is a logarithm of a quotient, a difference of roots or of arctangents, or $\frac{n}{(n+1)!}$ → telescoping: rewrite as b of n plus 1 minus b of n and look at the limit of b

Example: $\ln\frac{n}{n+1}$ : $s_N = -\ln(N+1)$ , diverges
If the statement is a repeating decimal or a process repeated with a fixed fraction → write it as a geometric series and sum it

Example: $0.\overline{36} = \frac{36/100}{1 - 1/100} = \frac{4}{11}$
If $a_n \to 0$ and none of the forms above → this chapter cannot decide; bound s_n by hand, or wait for the tests of the next chapters

Example: $\sum \frac{1}{\sqrt{n}}$ : $s_n \ge n \cdot \frac{1}{\sqrt{n}} = \sqrt{n}$ , diverges

The first branch costs five seconds and ends a third of exam questions. Never start partial fractions on a series whose terms tend to $\frac{2}{5}$ .

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Proving that a telescoping series converges, and finding its sum

When to use it: Any series asked with determine whether it converges and, if so, find its sum, whose term can be written as a difference

1 Rewrite $a_n$ as a difference, showing the partial fraction computation or the logarithm rule used.
2 Write $s_N$ with its first three brackets and its last two, so that the cancellation can be SEEN.
3 State the closed form of $s_N$ with every surviving term, at both ends.
4 Take $\lim_{N\to\infty} s_N$ and name the limits used ( $\frac{1}{N+1} \to 0$ , $\arctan(N+1) \to \frac{\pi}{2}$ ).
5 Conclude with the sentence: the series converges because its partial sums converge, and its sum is the limit.

Concluding sentence

“Since $s_N = \frac{1}{2}\left(\frac{3}{2} - \frac{1}{N+1} - \frac{1}{N+2}\right) \to \frac{3}{4}$ , the series converges and $\sum_{n=1}^{\infty} \frac{1}{n(n+2)} = \frac{3}{4}$ .”

The trap: Writing $\sum \frac{1}{2n} - \sum \frac{1}{2(n+2)}$ as two infinite series before cancelling, or asserting the cancellation without writing any bracket.

Marking: Typically 1 mark for the decomposition, 1 for s_N in closed form, 1 for the limit and the concluding sentence. A correct number without s_N earns very little.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The first partial sum

Compute $s_1 = a_1$ from the statement and from your closed form of $s_N$ . If they differ, the closed form is wrong, usually by a surviving term.

$\sum \frac{2n+1}{n^2(n+1)^2}$ : $a_1 = \frac{3}{4}$ and $s_1 = 1 - \frac{1}{4} = \frac{3}{4}$ .
The sign and the size of the sum

Positive terms: the sum is larger than $a_1$ and larger than every partial sum. Negative terms: negative sum. A negative sum of positive terms means the geometric formula was used with $|r| \ge 1$ .

$\sum_{n \ge 1} \left(\frac{2}{3}\right)^n$ : $s_3 = \frac{38}{27} \approx 1.41 < 2$ , consistent.
The ratio by quotient

Divide $a_2$ by $a_1$ and $a_3$ by $a_2$ : the two numbers must agree, and that number is $r$ .

$\frac{4^n}{5^{n-1}}$ : $a_1 = 4$ , $a_2 = \frac{16}{5}$ , $a_3 = \frac{64}{25}$ ; both quotients are $\frac{4}{5}$ .
Multiply a decimal back

For a repeating decimal, multiply your fraction's denominator by the decimal, or divide by hand for two cycles.

$\frac{4}{11}$ : $40 = 3 \times 11 + 7$ , $70 = 6 \times 11 + 4$ , so $0.3636\ldots$

The typical problem, taken apart

The bouncing ball: total distance and total time

A ball is dropped from a height of $5$ m. Each time it hits the ground it rebounds to $\frac{1}{4}$ of the height it fell from. A fall from height $h$ lasts $\sqrt{\frac{h}{5}}$ seconds (taking $g = 10$ m/s $^2$ ), and a rise to height $h$ takes as long.

Find the total distance travelled and the total time before the ball comes to rest. Exact answers.

Each arc is a quarter as high as the previous one and half as long: the heights form a geometric series of ratio

\frac{1}{4}

, the durations one of ratio

\frac{1}{2}

Step 1

Heights reached after the drop: $h_k = 5\left(\frac{1}{4}\right)^k$ , so $\frac{5}{4}$ , $\frac{5}{16}$ , $\frac{5}{64}$ , and so on, for $k \ge 1$ .

Why

Naming the sequence of heights first separates the ONE initial drop from the infinitely many bounces, each of which is travelled twice, up and down.

Step 2

Distance: $D = 5 + 2\sum_{k=1}^{\infty} 5\left(\frac{1}{4}\right)^k = 5 + 2 \cdot \frac{5/4}{1 - 1/4} = 5 + 2 \cdot \frac{5}{3} = \frac{25}{3}$ m.

Why

The series starts at $k = 1$ , so its first term is $\frac{5}{4}$ , not $5$ : this is where most copies lose the answer, writing $\frac{5}{1 - 1/4} = \frac{20}{3}$ for the bounces.

Step 3

Durations: the drop lasts $\sqrt{1} = 1$ s; bounce $k$ lasts $2\sqrt{\frac{h_k}{5}} = 2\sqrt{\frac{1}{4^k}} = \frac{2}{2^k}$ s.

Why

Taking the square root of a geometric sequence gives another geometric sequence, with ratio $\sqrt{\frac{1}{4}} = \frac{1}{2}$ : the time series has its own ratio, it is NOT the distance ratio.

Step 4

Time: $T = 1 + \sum_{k=1}^{\infty} \frac{2}{2^k} = 1 + 2 \cdot \frac{1/2}{1 - 1/2} = 1 + 2 = 3$ s.

Why

Both ratios are less than $1$ , so both series converge: infinitely many bounces fit in $3$ seconds. The sum is the limit of the partial times $2$ , $2.5$ , $2.75, \dots$ , visible on the figure.

Step 5

Check: after the drop and the first bounce, the ball has travelled $5 + \frac{5}{2} = 7.5$ m in $2$ s; the totals $\frac{25}{3} \approx 8.3$ m and $3$ s are larger, and not by much.

Why

A partial sum must stay below the total for positive terms, and the first terms carry most of the sum when the ratio is small.

The conclusion, written out

“The ball travels $\frac{25}{3}$ m in total and comes to rest after $3$ seconds, both sums being geometric series with ratios $\frac{1}{4}$ and $\frac{1}{2}$ , less than $1$ .”

The classic mistake on this problem: Counting every bounce once instead of twice, starting the bounce series at $5$ instead of $\frac{5}{4}$ , or using the ratio $\frac{1}{4}$ for the times, which gives $1 + \frac{1}{1 - 1/4} = \frac{7}{3}$ s instead of $3$ : the durations scale by $\frac{1}{2}$ , the square root of the height ratio.

Learn by heart

• $\sum a_n$ converges when $s_n = a_1 + \dots + a_n$ has a finite limit; that limit is the sum.
• Geometric: sum $= \frac{\text{first term}}{1 - r}$ if $|r| < 1$ ; diverges if $|r| \ge 1$ . First term by substitution of the starting index.
• Telescoping: $\sum (b_{n+1} - b_n) = \lim b_n - b_1$ , when the limit exists.
• $\lim a_n \ne 0$ : diverges. $\lim a_n = 0$ : no conclusion.
• $\sum \frac{1}{n}$ diverges ( $s_{2^k} \ge 1 + \frac{k}{2}$ ).
• Split a series only when every piece converges; otherwise work on $s_N$ .
• $0.\overline{d_1 \ldots d_p} = \frac{d_1 \ldots d_p}{10^p - 1}$ , and $0.\overline{9} = 1$ .

Frequently asked questions

How do I find the sum of a geometric series that starts at n = 1?

Compute the first term by substituting the starting index, then divide it by one minus the ratio, provided the absolute value of the ratio is less than one. For the series of two thirds to the power n starting at n equals 1, the first term is two thirds and the sum is 2, not 3. One over one minus r only works when the first term is 1.

If the terms of a series go to zero, does the series converge?

Not necessarily. Terms tending to zero is necessary for convergence but not sufficient. The harmonic series, the sum of one over n, has terms tending to zero and still diverges, because its partial sums grow without bound. The divergence test can only prove divergence; when the limit of the terms is zero, it tells you nothing.

How do I know how many terms survive in a telescoping series?

Write out the partial sum with its first three brackets and its last two. If each negative piece cancels with the positive piece of the next bracket, one term survives at each end; if it cancels two brackets later, two terms survive at each end. Never assume the cancellation: the written brackets are what the marker wants to see.

Can I split a series into two sums when finding its value?

Only when both resulting series converge. Splitting the sum of one over n minus one over n plus 1 into two harmonic series gives infinity minus infinity, which has no value. Work on the finite partial sum instead, where every rearrangement is allowed, and take the limit at the very end.

Why is 0.999 repeating equal to 1?

Because the decimal is the geometric series nine tenths plus nine hundredths plus nine thousandths and so on, with first term nine tenths and ratio one tenth. Its sum is nine tenths divided by nine tenths, which is exactly 1. Every partial sum is less than 1, but the decimal means the limit of those partial sums, and that limit is 1.

Practise it

Corrected exercises: Series: geometric, telescoping and the divergence test, MATH 141 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Sequences Next sheet Integral and comparison tests

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Revision sheet: geometric series, telescoping series and the divergence test (MATH 141)

The essentials

Two sequences in every question: the terms and the partial sums

The two series you can actually sum

The rules in table form

Which verdict, from which form of the series

The mistakes that cost marks

1. Using 1 over 1 minus r on a series that starts at n = 1

2. Reading the ratio off the letter under the exponent

3. Concluding convergence from terms that tend to 0

4. Taking the limit of the terms for the sum

5. Splitting a telescoping series into two divergent series

6. Keeping one surviving term when the gap is two

7. Inserting brackets into a divergent series

8. Using the geometric formula when the ratio is at least 1

Which method to choose

What to do with a series, by the FORM of its general term

How the answer is expected to be written

Proving that a telescoping series converges, and finding its sum

Check before you hand in

The typical problem, taken apart

The bouncing ball: total distance and total time

Learn by heart

Frequently asked questions

How do I find the sum of a geometric series that starts at n = 1?

If the terms of a series go to zero, does the series converge?

How do I know how many terms survive in a telescoping series?

Can I split a series into two sums when finding its value?

Why is 0.999 repeating equal to 1?

Corrected exercises: Series: geometric, telescoping and the divergence test, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?