MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: power series (MATH 141)

Q: How do I find the interval of convergence of a power series?

Apply the ratio test to the absolute value of the general term, with x fixed, and solve limit less than one: this gives the centre and the radius. Then substitute each endpoint into the series and test the two numerical series separately, with the divergence test, a p-series, a comparison or the alternating series test. The interval is written with each bracket decided by its endpoint.

Q: Why can't the ratio test decide the endpoints of a power series?

Because the radius is defined as the place where the ratio limit equals one, and a limit equal to one is exactly the case where the ratio test gives no conclusion. The same holds for the root test. At an endpoint the series becomes an ordinary numerical series, and only the tests that can decide it, such as p-series or the alternating series test, may be used.

Q: Does differentiating a power series change its interval of convergence?

It keeps the radius but not necessarily the endpoints. Differentiating multiplies each term by n, so a series that converged at an endpoint can diverge there after differentiation. Integrating divides by n plus one and can only gain an endpoint. The series of x to the n over n converges at minus one, its derivative does not.

Q: How do I write 1 over (a + x) as a power series?

Factor a out of the denominator to create the form one over one minus u: one over (a plus x) equals one over a, times one over (1 plus x over a). Then u is minus x over a, the series is the geometric series in u multiplied by one over a, and it converges when the absolute value of x is less than the absolute value of a. Check at x equal to zero: the series must give one over a.

This sheet is not a summary of sections 11.8 and 11.9 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on power series in MATH 141 at McGill University, and which precise gesture avoids each loss.

Every number below is done by hand, as on the exam, and every value quoted has been checked by an independent computation. The rule that runs through the whole sheet: the radius comes from one test, each endpoint from its own.

Updated on October 17, 2026

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The thread of the chapter

The ratio test finds the RADIUS, never the interval: each endpoint is a separate numerical series with its own test, every operation on a series (substitute, recentre, multiply by a power of x, differentiate, integrate) keeps track of R but not of the endpoints, and outside the interval the series is worth nothing, even where the function exists.

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

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Power series
Radius of convergence
Interval of convergence
Centre of a power series
Ratio test
Geometric series
Term-by-term differentiation
Term-by-term integration
Alternating series test
Abel's theorem

The essentials

Radius from one test, each endpoint from its own

• A power series $\sum c_n(x - a)^n$ converges absolutely for $|x - a| < R$ and diverges for $|x - a| > R$ . It always converges at its centre $a$ ; $R$ can be $0$ or $\infty$ .
• $R$ : ratio test on $|a_n|$ with $x$ fixed, $\left|\frac{a_{n+1}}{a_n}\right| \to \frac{|x - a|}{R}$ . When every factor is a power $n$ , the root test is faster.
• At $|x - a| = R$ the ratio limit is exactly $1$ : the test is silent BY CONSTRUCTION. Substitute $x = a - R$ and $x = a + R$ , write the two numerical series, test each one.
• Four possible answers: $(a - R, a + R)$ , $[a - R, a + R)$ , $(a - R, a + R]$ , $[a - R, a + R]$ . Only the endpoint tests choose between them.
• Read the centre after factoring: $(2x - 3)^n = 2^n\left(x - \frac{3}{2}\right)^n$ , centre $\frac{3}{2}$ , and $|2x - 3| < 1$ gives $R = \frac{1}{2}$ .

For

\sum \frac{(x - 1)^n}{\sqrt{n}\,2^n}

the ratio test gives only

(-1, 3)

; the alternating series test closes the bracket at

-1

, the

p

-series

\sum \frac{1}{\sqrt{n}}

leaves it open at

3

The midpoint of the interval must be the centre, and its half-length the radius: a ten-second check that catches every misread centre.

Everything comes from 1/(1 - u)

• $\frac{1}{1 - u} = \sum_{n=0}^\infty u^n$ for $|u| < 1$ , and never at $|u| = 1$ . Put the function in the form $\frac{\text{constant}}{1 - u}$ and read $u$ : the interval is $|u| < 1$ .
• Create the $1$ by factoring: $\frac{1}{2 + x} = \frac{1}{2} \cdot \frac{1}{1 + x/2} = \sum \frac{(-1)^n x^n}{2^{n+1}}$ , $|x| < 2$ .
• Recentre at $a$ by rewriting around $a$ : $\frac{1}{x + 5} = \frac{1}{7 + (x - 2)} = \sum \frac{(-1)^n (x - 2)^n}{7^{n+1}}$ , $|x - 2| < 7$ .
• Differentiate or integrate term by term inside $(a - R, a + R)$ : same $R$ , endpoints re-tested, and a constant $C$ found at the centre.
• $\ln(1 + x) = \sum_{n\ge1} (-1)^{n+1}\frac{x^n}{n}$ on $(-1, 1]$ , $\arctan x = \sum_{n\ge0} (-1)^n\frac{x^{2n+1}}{2n + 1}$ on $[-1, 1]$ , $\frac{1}{(1 - x)^2} = \sum_{n\ge0} (n + 1)x^n$ on $(-1, 1)$ .

(-1, 1)

the partial sums

S_1

S_3

S_6

\sum x^n

close in on

\frac{1}{1 - x}

; outside the dashed lines

x = \pm 1

they run away, whatever the function does.

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.

What each operation does to the radius and to the endpoints

Start from a series of radius $R$ . The left column is the operation, the middle one what happens to $R$ , the right one what the endpoints need afterwards. The red line is the only one that produces no value at all.

Operation	Radius	Endpoints after it
replace $x$ by $cx$	$\frac{R}{\|c\|}$	re-test
Example: $\frac{1}{1 + 3x} = \sum (-3)^n x^n$ : $R = \frac{1}{3}$ , interval $\left(-\frac{1}{3}, \frac{1}{3}\right)$ .
replace $x$ by $x^2$	$\sqrt{R}$	re-test
Example: $\sum \frac{u^n}{2^n}$ has $R = 2$ , so $\sum \frac{x^{2n}}{2^n}$ has $R = \sqrt{2}$ .
multiply by $x^k$	unchanged	unchanged
Example: $\frac{x^2}{4 - x} = \sum \frac{x^{n+2}}{4^{n+1}}$ on $(-4, 4)$ , like $\frac{1}{4 - x}$ .
differentiate	unchanged	may LOSE an endpoint
Example: $\sum \frac{x^n}{n^2}$ on $[-1, 1]$ gives $\sum \frac{x^{n-1}}{n}$ on $[-1, 1)$ .
integrate	unchanged	may GAIN an endpoint
Example: $\frac{1}{1 + x^2}$ on $(-1, 1)$ gives $\arctan x$ on $[-1, 1]$ .
evaluate at $\|x - a\| > R$	not applicable	the series diverges no value
Example: $\ln(1 + x)$ at $x = 2$ : the terms $\frac{2^n}{n}$ tend to $\infty$ , so the series is not $\ln 3$ . Same form, other result: $\frac{1}{1 - x}$ at $x = -\frac{1}{2}$ : $\sum \left(-\frac{1}{2}\right)^n = \frac{2}{3}$ , correct; at $x = -2$ : $\sum (-2)^n$ diverges while $\frac{1}{1 - (-2)} = \frac{1}{3}$ . What to do: Find another series, or another point, inside its interval: $\ln 3 = -\ln\left(1 - \frac{2}{3}\right) = \sum \frac{(2/3)^n}{n}$ .

The radius column is a free result from the theorem or from $|u| < 1$ ; the endpoint column is always work to do on the copy.

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. Giving the open interval as the answer

2 marks, the two endpoint marks of the question

What not to write

“ $\left|\frac{a_{n+1}}{a_n}\right| \to \frac{|x|}{3} < 1$ , so the interval of convergence of $\sum \frac{x^n}{n\,3^n}$ is $(-3, 3)$ .”

What to write

“ $R = 3$ . At $x = 3$ : $\sum \frac{1}{n}$ diverges. At $x = -3$ : $\sum \frac{(-1)^n}{n}$ converges by the alternating series test. Interval $[-3, 3)$ .”

The ratio test gives the open interval

(-3, 3)

. Each endpoint is then tested on its own: closed at

-3

(green), open at

3

(red), so the interval is

[-3, 3)

Why: The ratio test decides only strictly inside and strictly outside. The endpoints are two more series, and the answer is not complete until each has a verdict with its test named.

2. Testing an endpoint with the ratio test

1 mark per endpoint

What not to write

“At $x = 3$ , $\lim\left|\frac{a_{n+1}}{a_n}\right| = 1$ , so the series diverges there.”

What to write

“At $x = 3$ the ratio limit is $1$ , which decides nothing. The series is $\sum \frac{1}{n}$ , a $p$ -series with $p = 1$ : it diverges.”

Why: $L = 1$ is the silent case of the ratio test, and at an endpoint it is $1$ by construction. The conclusion must come from a test that CAN decide: divergence test, $p$ -series, comparison, alternating series test.

3. Misreading the centre or the radius

the whole question, since every endpoint is then wrong

What not to write

“ $\sum \frac{(2x - 3)^n}{n}$ is centred at $3$ with radius $1$ .”

What to write

“ $(2x - 3)^n = 2^n\left(x - \frac{3}{2}\right)^n$ : centre $\frac{3}{2}$ , and $|2x - 3| < 1 \iff \left|x - \frac{3}{2}\right| < \frac{1}{2}$ , so $R = \frac{1}{2}$ and the interval is $[1, 2)$ .”

(2x - 3)^n = 2^n\left(x - \frac32\right)^n

: centre

\frac32

, radius

\frac12

, interval

[1, 2)

(green). Reading the centre

3

and radius

1

off

(2x - 3)

(red) misplaces both.

Why: The centre is the point where the base of the power vanishes, and the radius is measured in $x$ , not in $2x$ . The check: the midpoint of $[1, 2)$ is $\frac{3}{2}$ and its half-length $\frac{1}{2}$ .

4. Evaluating a series outside its interval

the whole question

What not to write

“ $\frac{1}{1 - x} = \sum x^n$ , so at $x = -1.5$ the series gives $\frac{1}{2.5} = 0.4$ .”

What to write

“The identity holds only for $|x| < 1$ . At $x = -1.5$ the terms $(-1.5)^n$ grow without bound: the series diverges, although $\frac{1}{1 - x} = 0.4$ exists.”

Left of the dashed line

x = -1

S_4

and

S_5

leave

\frac{1}{1 - x}

in opposite directions, although the function is perfectly smooth there.

Why: A power series knows its function only inside its interval. Beyond it the partial sums swing further and further from the function, and no value can be read off them.

5. Forgetting the constant after integrating

1 to 2 marks

What not to write

“ $\frac{d}{dx}\ln(3 + x) = \frac{1}{3 + x} = \sum \frac{(-1)^n x^n}{3^{n+1}}$ , so $\ln(3 + x) = \sum \frac{(-1)^n x^{n+1}}{(n + 1)3^{n+1}}$ .”

What to write

“ $\ln(3 + x) = C + \sum \frac{(-1)^n x^{n+1}}{(n + 1)3^{n+1}}$ , and at $x = 0$ : $\ln 3 = C$ . So $\ln(3 + x) = \ln 3 + \sum_{n\ge0} \frac{(-1)^n x^{n+1}}{(n + 1)3^{n+1}}$ on $(-3, 3]$ .”

Why: Term-by-term integration gives an antiderivative, so a constant. At the centre the series is worth $0$ and the function $\ln 3$ : the check at $x = 0$ finds $C$ in one line.

6. Differentiating a number instead of a function

the whole question

What not to write

“ $\sum_{n\ge0} \left(\frac{1}{3}\right)^n = \frac{3}{2}$ , and the derivative of $\frac{3}{2}$ is $0$ , so $\sum_{n\ge1} n\left(\frac{1}{3}\right)^{n-1} = 0$ .”

What to write

“On $(-1, 1)$ , $\sum_{n\ge1} n x^{n-1} = \frac{1}{(1 - x)^2}$ ; at $x = \frac{1}{3}$ this gives $\frac{9}{4}$ , so $\sum \frac{n}{3^n} = \frac{1}{3} \cdot \frac{9}{4} = \frac{3}{4}$ .”

Why: Differentiation acts on functions of $x$ . Keep $x$ a variable until the identity between functions is established, and substitute the number on the last line only.

7. Assuming a derived series keeps its endpoints

1 mark per endpoint

What not to write

“ $\sum \frac{x^n}{n}$ converges on $[-1, 1)$ , so its derivative $\sum x^{n-1}$ converges on $[-1, 1)$ too, by the theorem.”

What to write

“The theorem gives the same radius $1$ . At $x = -1$ , $\sum (-1)^{n-1}$ has terms that do not tend to $0$ : it diverges. The derivative converges on $(-1, 1)$ .”

Same radius

1

on all three lines, three different brackets: the derivative loses

x = -1

, the antiderivative gains

x = 1

Why: Differentiating multiplies the $n$ th term by $n$ , which can only make things worse at an endpoint; integrating divides by $n + 1$ , which can only help. The theorem speaks about $R$ , never about the brackets.

8. A coefficient formula on a series with missing powers

the whole question

What not to write

“ $\sum \frac{x^{2n}}{4^n}$ has $c_n = \frac{1}{4^n}$ , so $R = \lim\left|\frac{c_n}{c_{n+1}}\right| = 4$ .”

What to write

“Ratio test on the terms: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{x^2}{4} < 1 \iff |x| < 2$ . So $R = 2$ , and at $x = \pm 2$ every term is $1$ : interval $(-2, 2)$ .”

Why: $\frac{1}{4^n}$ is the coefficient of $x^{2n}$ , not of $x^n$ : the odd coefficients are $0$ and the shortcut compares the wrong numbers. Always apply the test to the terms as written.

Which method to choose

Which test at an endpoint, by the FORM of the series

The ratio test has given $|x - a| < R$ . Substitute $x = a - R$ and $x = a + R$ , write each numerical series in full, and read its form

If the terms do not tend to 0 (they are all plus or minus 1, or they grow like n) → divergence test: the endpoint is excluded in one line

Example: $\sum \frac{n x^n}{4^n}$ at $x = 4$ : the terms are $n$ , diverges
If the absolute values behave like $\frac{1}{n^p}$ with $p > 1$ → absolute convergence by p-series or comparison, which settles BOTH endpoints at once

Example: $\sum \frac{x^n}{n^2\,5^n}$ at $x = \pm 5$ : $\frac{1}{n^2}$ , both included
If constant sign, absolute values like $\frac{1}{n^p}$ with $p \le 1$ → p-series or limit comparison: diverges

Example: $\sum \frac{(-1)^n x^n}{\sqrt{n + 1}}$ at $x = -1$ : $\sum \frac{1}{\sqrt{n + 1}}$ diverges
If alternating sign, absolute values decreasing to $0$ but like $\frac{1}{n^p}$ with $p \le 1$ → alternating series test: converges, conditionally

Example: $\sum \frac{x^n}{n\,3^n}$ at $x = -3$ : $\sum \frac{(-1)^n}{n}$ converges
If the series came from $\frac{1}{1 - u}$ with nothing else done to it → $|u| = 1$ at both endpoints: both excluded, no further test

Example: $\frac{1}{1 + 3x}$ at $x = \pm\frac{1}{3}$ : terms $\pm 1$
If the term is a power $n$ of a base that tends to $1$ → form $1^\infty$ : compute the limit of the terms with a logarithm before concluding

Example: $\left(1 - \frac{5}{2n + 6}\right)^n \to e^{-5/2} \ne 0$ : diverges

Never the ratio or root test at an endpoint: there its limit is $1$ by construction, the only value on which it is silent.

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.

Writing an interval of convergence

When to use it: Any question that says find the radius and the interval of convergence

1 Name the term $a_n$ , with $x$ fixed, and apply the ratio test (or the root test) to $|a_n|$ , with the limit written.
2 Solve limit $< 1$ in the form $|x - a| < R$ : state the centre and the radius separately.
3 Substitute each endpoint, write the numerical series in full, name its test and verify the hypotheses of that test.
4 Conclude with the interval, each bracket justified by the verdict of its endpoint.

Concluding sentence

“By the ratio test the series converges absolutely for $|x| < 3$ and diverges for $|x| > 3$ , so $R = 3$ . At $x = 3$ it is $\sum \frac{1}{n}$ , which diverges; at $x = -3$ it is $\sum \frac{(-1)^n}{n}$ , which converges by the alternating series test. The interval of convergence is $[-3, 3)$ .”

The trap: Writing the two endpoint verdicts without the series they come from: the marker cannot tell which test applies to which series, and gives nothing for a verdict alone.

Marking: Typically 2 marks for the radius with the ratio test written, 1 to 2 marks per endpoint (series written, test named, hypotheses checked), and the interval with the right brackets.

Check before you hand in

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

The centre test

Substitute the centre into the series you built: only its first term survives, and it must equal the value of the function there.

$\frac{1}{2 + x}$ at $x = 0$ is $\frac{1}{2}$ ; a series starting with $1$ has lost a factor $\frac{1}{2}$ .
The midpoint test

Whatever the brackets, the interval found must have the centre as its midpoint and the radius as its half-length.

$\sum \frac{(2x - 3)^n}{n}$ : $[1, 2)$ has midpoint $\frac{3}{2}$ and half-length $\frac{1}{2}$ , as it should.
The asymptote test

For a rational function built from 1/(1 - u), the radius is the distance from the centre to the nearest point where the function blows up. Use it as a check, not as a proof.

$\frac{1}{x + 5}$ about $a = 2$ : the distance from $2$ to $-5$ is $7$ , the radius found.
The first-terms test

When a numerical sum is obtained by evaluating a series, add three or four terms by hand: the value must be close.

$\sum \frac{n}{2^n} = 2$ : $\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} = \frac{13}{8}$ , and the terms left are small and positive.

The typical problem, taken apart

Interval, closed form and a numerical sum

Let $g(x) = \sum_{n=1}^\infty \frac{(x - 1)^n}{n\,2^n}$ . Find the interval of convergence, a closed form for $g$ , and deduce the sum of $\sum_{n=1}^\infty \frac{(-1)^n}{n\,2^n}$ .

No calculator. Every step must be justified as on a MATH 141 final.

Step 1

Ratio test, $x$ fixed: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x - 1|}{2} \cdot \frac{n}{n + 1} \to \frac{|x - 1|}{2}$ , which is $< 1$ iff $|x - 1| < 2$ . Centre $1$ , $R = 2$ , open interval $(-1, 3)$ .

Why

The limit written in full is the first mark; the centre and the radius are stated separately because the next two steps use them.

Step 2

At $x = 3$ : $\sum \frac{2^n}{n\,2^n} = \sum \frac{1}{n}$ , diverges ( $p = 1$ ). At $x = -1$ : $\sum \frac{(-2)^n}{n\,2^n} = \sum \frac{(-1)^n}{n}$ , alternating with $\frac{1}{n}$ decreasing to $0$ : converges. Interval $[-1, 3)$ .

Why

Each endpoint is its own numerical series with its own test; the ratio test gives $1$ at both and could decide neither.

Step 3

For $|x - 1| < 2$ , differentiate term by term: $g'(x) = \sum_{n\ge1} \frac{(x - 1)^{n-1}}{2^n} = \frac{1}{2} \cdot \frac{1}{1 - \frac{x - 1}{2}} = \frac{1}{3 - x}$ .

Why

Differentiating removes the $n$ of the denominator and leaves a geometric series with $u = \frac{x - 1}{2}$ , whose sum is known.

Step 4

Integrate: $g(x) = -\ln(3 - x) + C$ . At the centre, $g(1) = 0$ (every term contains $(x - 1)^n$ with $n \ge 1$ ), so $0 = -\ln 2 + C$ , $C = \ln 2$ and $g(x) = \ln\frac{2}{3 - x}$ .

Why

The constant is fixed at the centre, where the series is worth its constant term; without it, $g(1)$ would be $-\ln 2$ , not $0$ .

Step 5

$x = 0$ lies in $[-1, 3)$ and gives $(x - 1)^n = (-1)^n$ : $\sum \frac{(-1)^n}{n\,2^n} = g(0) = \ln\frac{2}{3}$ . Check: $-\frac{1}{2} + \frac{1}{8} - \frac{1}{24} + \frac{1}{64} \approx -0.40$ , and $\ln 2 - \ln 3 \approx 0.69 - 1.10 = -0.41$ .

Why

A point is substituted only once it is checked to be inside the interval; the first-terms test then confirms the value.

The conclusion, written out

“The series converges exactly on $[-1, 3)$ , where $g(x) = \ln\frac{2}{3 - x}$ ; hence $\sum_{n=1}^\infty \frac{(-1)^n}{n\,2^n} = \ln\frac{2}{3}$ .”

The classic mistake on this problem: Stopping at $(-1, 3)$ , or writing $g(x) = -\ln(3 - x)$ without the constant, which gives $\sum \frac{(-1)^n}{n\,2^n} = -\ln 3$ : negative, plausible, and wrong, since the first-terms test gives about $-0.40$ , not $-1.10$ .

Learn by heart

• The ratio test gives $R$ , never the interval: at $|x - a| = R$ its limit is $1$ and it is silent.
• Each endpoint: write the numerical series, name its test, check its hypotheses.
• $\frac{1}{1 - u} = \sum u^n$ for $|u| < 1$ only; never at $|u| = 1$ .
• Factor to create the $1$ : $\frac{1}{2 + x} = \frac{1}{2} \cdot \frac{1}{1 + x/2}$ , and $|x| < 2$ .
• Substituting $x^2$ takes the square root of $R$ ; substituting $cx$ divides it by $|c|$ .
• Differentiating and integrating keep $R$ ; differentiating may lose an endpoint, integrating may gain one.
• After integrating, find $C$ at the centre. After differentiating, evaluate on the last line only.
• Outside its interval a series is worth nothing, even where the function exists.

Frequently asked questions

How do I find the interval of convergence of a power series?

Apply the ratio test to the absolute value of the general term, with x fixed, and solve limit less than one: this gives the centre and the radius. Then substitute each endpoint into the series and test the two numerical series separately, with the divergence test, a p-series, a comparison or the alternating series test. The interval is written with each bracket decided by its endpoint.

Why can't the ratio test decide the endpoints of a power series?

Because the radius is defined as the place where the ratio limit equals one, and a limit equal to one is exactly the case where the ratio test gives no conclusion. The same holds for the root test. At an endpoint the series becomes an ordinary numerical series, and only the tests that can decide it, such as p-series or the alternating series test, may be used.

Does differentiating a power series change its interval of convergence?

It keeps the radius but not necessarily the endpoints. Differentiating multiplies each term by n, so a series that converged at an endpoint can diverge there after differentiation. Integrating divides by n plus one and can only gain an endpoint. The series of x to the n over n converges at minus one, its derivative does not.

How do I write 1 over (a + x) as a power series?

Factor a out of the denominator to create the form one over one minus u: one over (a plus x) equals one over a, times one over (1 plus x over a). Then u is minus x over a, the series is the geometric series in u multiplied by one over a, and it converges when the absolute value of x is less than the absolute value of a. Check at x equal to zero: the series must give one over a.

Practise it

Corrected exercises: Power series, 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

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Ratio and root tests Next sheet Taylor and Maclaurin series

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math141-power-series. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: power series (MATH 141)

The essentials

Radius from one test, each endpoint from its own

Everything comes from 1/(1 - u)

The rules in table form

What each operation does to the radius and to the endpoints

The mistakes that cost marks

1. Giving the open interval as the answer

2. Testing an endpoint with the ratio test

3. Misreading the centre or the radius

4. Evaluating a series outside its interval

5. Forgetting the constant after integrating

6. Differentiating a number instead of a function

7. Assuming a derived series keeps its endpoints

8. A coefficient formula on a series with missing powers

Which method to choose

Which test at an endpoint, by the FORM of the series

How the answer is expected to be written

Writing an interval of convergence

Check before you hand in

The typical problem, taken apart

Interval, closed form and a numerical sum

Learn by heart

Frequently asked questions

How do I find the interval of convergence of a power series?

Why can't the ratio test decide the endpoints of a power series?

Does differentiating a power series change its interval of convergence?

How do I write 1 over (a + x) as a power series?

Corrected exercises: Power series, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?