Sequences and Series | AP - Wyatt's Notes

Sequences (CED BC Unit 10)

A sequence is an ordered list of numbers: $a_1, a_2, a_3, \ldots$ . Formally, a sequence is a Function from the positive integers (or a subset thereof) to the real numbers: $a \colon \mathbb{N} \to \mathbb{R}$ Written as $\{a_n\}_{n=1}^{\infty}$ or $\{a_n\}$ .

A sequence $\{a_n\}$ converges to a limit $L$ if:

\lim_{n \to \infty} a_n = L

This means: for every $\epsilon \gt 0$ There exists an integer $N$ such that $|a_n - L| \lt \epsilon$ for all $n \ge N$ . The terms eventually get and stay arbitrarily close to $L$ .

If no such limit exists, the sequence diverges.

Bounded and Monotone Sequences

Bounded above: $a_n \le M$ for all $n$ and some $M$ .
Bounded below: $a_n \ge m$ for all $n$ and some $m$ .
Bounded: bounded both above and below.
Monotone increasing: $a_{n+1} \ge a_n$ for all $n$ .
Monotone decreasing: $a_{n+1} \le a_n$ for all $n$ .
Eventually monotone: the monotonicity holds for all $n$ beyond some index $N$ .

Monotone Convergence Theorem. Every bounded monotone sequence converges. This is one of the most Powerful existence theorems in analysis: it guarantees convergence without requiring you to find the Limit explicitly.

Corollary: A monotone increasing sequence that is not bounded above diverges to $+\infty$ . A Monotone decreasing sequence that is not bounded below diverges to $-\infty$ .

Common Sequences

| Sequence | Convergence | Limit | | -------------------------------------- | ----------- | ------- | --------- | --- | | $a_n = \frac{1}{n}$ | Converges | $0$ | | $a_n = r^n$ ( $| r | \lt 1$ ) | Converges | $0$ | | $a_n = r^n$ ( $| r | \ge 1$ ) | Diverges | — | | $a_n = \left(1 + \frac{1}{n}\right)^n$ | Converges | $e$ | | $a_n = \frac{n!}{n^n}$ | Converges | $0$ | | $a_n = (-1)^n$ | Diverges | — | | $a_n = \sqrt{n+1} - \sqrt{n}$ | Converges | $0$ | | $a_n = \frac{\ln n}{n}$ | Converges | $0$ |

Proof: $\frac{n!}{n^n} \to 0$

Write out the terms:

0 \lt \frac{n!}{n^n} = \frac{1 \cdot 2 \cdot 3 \cdots n}{n \cdot n \cdot n \cdots n} = \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{n}{n}

The first $\lfloor n/2 \rfloor$ factors are each at most $\frac{1}{2}$ So:

0 \lt \frac{n!}{n^n} \le \left(\frac{1}{2}\right)^{\lfloor n/2 \rfloor} \to 0

By the squeeze theorem, $\frac{n!}{n^n} \to 0$ .

Proof: $\frac{\ln n}{n} \to 0$

Since $\ln n$ grows slower than any positive power of $n$ We have $\ln n \lt \sqrt{n}$ for Sufficiently large $n$ . Therefore $0 \lt \frac{\ln n}{n} \lt \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}$ And $\frac{1}{\sqrt{n}} \to 0$ So by the squeeze theorem, $\frac{\ln n}{n} \to 0$ .

Series (CED BC Unit 10)

An infinite series is the sum of the terms of an infinite sequence:

\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots

Partial Sums

The $n$ Th partial sum is $S_n = \sum_{k=1}^{n} a_k$ . The series converges if and only if the Sequence of partial sums $\{S_n\}$ converges:

\sum_{n=1}^{\infty} a_n = L \iff \lim_{n \to \infty} S_n = L

If $\{S_n\}$ diverges, the series diverges.

The $n$ Th-Term Test (Divergence Test)

If $\displaystyle\lim_{n \to \infty} a_n \ne 0$ Then $\displaystyle\sum a_n$ diverges.

Proof (by contrapositive): If $\sum a_n$ converges to $L$ Then $S_n \to L$ and $S_{n-1} \to L$ . Since $a_n = S_n - S_{n-1}$ We get $a_n \to L - L = 0$ .

Caution: If $\displaystyle\lim_{n \to \infty} a_n = 0$ The test is inconclusive. The series may Converge or diverge. The harmonic series $\sum \frac{1}{n}$ is the canonical counterexample.

Example Does $\displaystyle\sum_{n=1}^{\infty} \frac{n}{n+1}$ converge?

\lim_{n \to \infty} \frac{n}{n+1} = 1 \ne 0

By the $n$ Th-term test, the series diverges.

The Harmonic Series

\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots

Even though $\frac{1}{n} \to 0$ This series diverges. The proof groups terms:

1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \cdots

Each group exceeds $\frac{1}{2}$ : $\frac{1}{3} + \frac{1}{4} \gt \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ And so on. Since we can Form infinitely many groups each exceeding $\frac{1}{2}$ The partial sums diverge to $+\infty$ .

Geometric Series (CED BC Unit 10.2)

\sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r}, \quad |r| \lt 1

The series diverges when $|r| \ge 1$ .

Derivation. The $n$ Th partial sum is $S_n = a + ar + ar^2 + \cdots + ar^{n-1}$ . Then:

RS_n = ar + ar^2 + \cdots + ar^n

S_n - rS_n = a - ar^n \implies S_n(1 - r) = a(1 - r^n)

S_n = \frac{a(1 - r^n)}{1 - r}

When $|r| \lt 1$ , $r^n \to 0$ So $S_n \to \frac{a}{1 - r}$ .

Example Evaluate $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ .

Use partial fractions: $\displaystyle\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ .

S_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1}

\lim_{n \to \infty} S_n = 1

Worked Example: Telescoping with Partial Fractions

Evaluate $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2 - 1}$ .

Factor: $\frac{1}{n^2-1} = \frac{1}{(n-1)(n+1)}$ . By partial fractions: $\frac{1}{(n-1)(n+1)} = \frac{1}{2}\!\left(\frac{1}{n-1} - \frac{1}{n+1}\right)$ .

S_N = \frac{1}{2}\left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{N-1} - \frac{1}{N+1}\right)\right]

Most terms cancel, leaving:

S_N = \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{N} - \frac{1}{N+1}\right)

\lim_{N \to \infty} S_N = \frac{1}{2}\left(\frac{3}{2}\right) = \frac{3}{4}

The Integral Test (CED BC Unit 10.3)

If $f$ is continuous, positive, and decreasing on $[1, \infty)$ And $a_n = f(n)$ Then:

\sum_{n=1}^{\infty} a_n \quad \mathrm{and \quad \int_1^{\infty} f(x)\, dx

Either both converge or both diverge.

Why it works. The sum $\sum_{n=2}^{\infty} f(n)$ can be bounded by the integral:

\int_1^{\infty} f(x)\, dx \le \sum_{n=1}^{\infty} f(n) \le f(1) + \int_1^{\infty} f(x)\, dx

So if the integral converges, the sum is bounded above and (since terms are positive) must converge. If the integral diverges, the sum exceeds any bound and must diverge.

Remainder Estimate for the Integral Test

If $\sum a_n$ converges by the integral test and $R_n = S - S_n$ is the remainder after $n$ terms:

\int_{n+1}^{\infty} f(x)\, dx \le R_n \le \int_n^{\infty} f(x)\, dx

$p$ -Series

\sum_{n=1}^{\infty} \frac{1}{n^p}

Converges if $p \gt 1$
Diverges if $p \le 1$

This follows directly from the integral test: $\int_1^{\infty} \frac{dx}{x^p}$ converges if and only If $p \gt 1$ .

The $p$ -series with $p = 1$ is the harmonic series, which diverges. This is the “boundary case” that Separates convergence from divergence.

Comparison Tests (CED BC Unit 10.5)

Direct Comparison Test

Suppose $0 \le a_n \le b_n$ for all $n$ (eventually):

If $\sum b_n$ converges, then $\sum a_n$ converges.
If $\sum a_n$ diverges, then $\sum b_n$ diverges.

Intuition: If a larger sum converges, the smaller one must too. If a smaller sum diverges, the Larger one must too.

Limit Comparison Test

Suppose $a_n \gt 0$ and $b_n \gt 0$ for all $n$ And:

L = \lim_{n \to \infty} \frac{a_n}{b_n}

If $0 \lt L \lt \infty$ Then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.
If $L = 0$ and $\sum b_n$ converges, then $\sum a_n$ converges.
If $L = \infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.

The case $0 \lt L \lt \infty$ is the most commonly used: it says the two series have the “same order Of magnitude,” so they share the same convergence behavior.

Example Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ converges.

Compare with $\displaystyle\sum \frac{1}{n^2}$ (a convergent $p$ -series with $p = 2$ ):

\lim_{n \to \infty} \frac{1/(n^2+1)}{1/n^2} = \lim_{n \to \infty} \frac{n^2}{n^2 + 1} = 1

Since $0 \lt 1 \lt \infty$ Both series converge by the limit comparison test.

Worked Example: Choosing the Right Comparison

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{3n + 1}{n^3 - 2}$ converges.

For large $n$ , $\frac{3n+1}{n^3-2} \approx \frac{3n}{n^3} = \frac{3}{n^2}$ .

Compare with $\sum \frac{1}{n^2}$ :

\lim_{n \to \infty} \frac{(3n+1)/(n^3-2)}{1/n^2} = \lim_{n \to \infty} \frac{n^2(3n+1)}{n^3-2} = \lim_{n \to \infty} \frac{3n^3 + n^2}{n^3 - 2} = 3

Since $L = 3 \in (0, \infty)$ Both series converge.

The Ratio Test (CED BC Unit 10.7)

For $\displaystyle\sum a_n$ Compute:

L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|

If $L \lt 1$ : the series converges absolutely.
If $L \gt 1$ (or $L = \infty$ ): the series diverges.
If $L = 1$ : the test is inconclusive.

The ratio test is especially useful when the terms involve factorials or exponentials, because the Ratio tends to simplify dramatically.

Why it connects to geometric series. If $\left|\frac{a_{n+1}}{a_n}\right| \to L \lt 1$ Then for Large $n$ the terms behave like a geometric series with ratio $L$ And geometric series converge When the ratio is less than 1.

Example Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{2^n}{n!}$ converges.

L = \lim_{n \to \infty} \frac{2^{n+1} / (n+1)!}{2^n / n!} = \lim_{n \to \infty} \frac{2}{n+1} = 0

Since $L = 0 \lt 1$ The series converges absolutely.

The Alternating Series Test (Leibniz Test)

If $\{a_n\}$ is a sequence that satisfies:

$a_n \gt 0$ for all $n$ (eventually),
$a_{n+1} \le a_n$ for all $n$ (eventually) — the terms decrease, and
$\displaystyle\lim_{n \to \infty} a_n = 0$

Then the alternating series $\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1} a_n$ converges.

Intuition: The partial sums oscillate, but the oscillations shrink because the terms decrease. The odd-indexed partial sums $S_1, S_3, S_5, \ldots$ form a decreasing sequence bounded below, and The even-indexed partial sums $S_2, S_4, S_6, \ldots$ form an increasing sequence bounded above. Both converge to the same limit.

Alternating Series Estimation Theorem

If $S = \displaystyle\sum_{n=1}^{\infty} (-1)^{n-1} a_n$ is a convergent alternating series, then The error in using $S_n$ to approximate $S$ satisfies:

|R_n| = |S - S_n| \le a_{n+1}

That is, the error is bounded by the first omitted term. This is remarkably useful: you can control The error by counting terms.

Example How many terms of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$ are needed to approximate The sum with error less than $0.001$ ?

The terms are $a_n = \frac{1}{n}$ Which decrease and approach 0.

We need $a_{n+1} = \frac{1}{n+1} \lt 0.001$ So $n + 1 \gt 1000$ Meaning $n \ge 1000$ .

At least 1000 terms are needed.

Absolute and Conditional Convergence

Absolutely convergent: $\displaystyle\sum |a_n|$ converges.
Conditionally convergent: $\displaystyle\sum a_n$ converges but $\displaystyle\sum |a_n|$ diverges.

Theorem. If a series converges absolutely, it converges.

Proof sketch. $-|a_n| \le a_n \le |a_n|$ So $0 \le a_n + |a_n| \le 2|a_n|$ . Since $\sum 2|a_n|$ Converges, $\sum (a_n + |a_n|)$ converges by the comparison test. Therefore $\sum a_n = \sum (a_n + |a_n|) - \sum |a_n|$ converges as the difference of two convergent series.

Riemann rearrangement theorem. A conditionally convergent series can be rearranged to converge To any real number, or to diverge. This is not true for absolutely convergent series, whose sum is Invariant under rearrangement.

Example Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ .

The alternating series converges by the alternating series test.

Check absolute convergence: $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$ is the harmonic series, Which diverges.

Therefore, the series converges conditionally.

Power Series (CED BC Unit 10.8)

A power series centered at $a$ is:

\sum_{n=0}^{\infty} c_n(x - a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots

A power series is a “polynomial with infinitely many terms.” The central question for any power Series is: for which values of $x$ does it converge?

Interval and Radius of Convergence

Every power series converges in an interval $(a - R, a + R)$ where $R$ is the radius of Convergence:

Use the ratio test to find $R$ : $\displaystyle R = \frac{1}{\lim_{n \to \infty} |c_{n+1}/c_n|}$ .
Check the endpoints separately (the ratio test is inconclusive when $L = 1$ ).

Case	Interval of Convergence
$R = 0$	Single point $\{a\}$
$R = \infty$	$(-\infty, \infty)$
$0 \lt R \lt \infty$	Check endpoints of $(a - R, a + R)$

Example Find the interval of convergence for $\displaystyle\sum_{n=1}^{\infty} \frac{(x-1)^n}{n}$ .

L = \lim_{n \to \infty} \left|\frac{(x-1)^{n+1}/(n+1)}{(x-1)^n/n}\right| = |x - 1| \cdot \lim_{n \to \infty} \frac{n}{n+1} = |x - 1|

Converges when $|x - 1| \lt 1$ I.e., $0 \lt x \lt 2$ . Radius $R = 1$ .

Check $x = 0$ : $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ converges (alternating Series).

Check $x = 2$ : $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$ diverges (harmonic series).

Interval of convergence: $[0, 2)$ .

Term-by-Term Differentiation and Integration

If $\displaystyle f(x) = \sum_{n=0}^{\infty} c_n(x-a)^n$ with radius $R$ Then:

F"(x) = \sum_{n=1}^{\infty} n c_n (x-a)^{n-1}, \quad \mathrm{same radius R

\int f(x)\, dx = C + \sum_{n=0}^{\infty} \frac{c_n (x-a)^{n+1}}{n+1}, \quad \mathrm{same radius R

Differentiation and integration of power series do not change the radius of convergence (though the Behaviour at the endpoints may change).

Taylor and Maclaurin Series (CED BC Unit 10.11)

The Taylor series of $f$ centered at $a$ is:

F(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n

When $a = 0$ This is called a Maclaurin series.

Why Taylor Series Work

The Taylor polynomial $T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$ is the unique Polynomial of degree $\le n$ whose value and first $n$ derivatives at $x = a$ match those of $f$ . As $n \to \infty$ If the remainder $R_n(x) = f(x) - T_n(x) \to 0$ Then the Taylor series converges to $f(x)$ .

Common Maclaurin Series

E^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots, \quad R = \infty

\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots, \quad R = \infty

\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots, \quad R = \infty

\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots, \quad R = 1

\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots, \quad R = 1

(1+x)^k = \sum_{n=0}^{\infty} \binom{k}{n} x^n, \quad R = 1

\arctan x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots, \quad R = 1

Taylor”s Inequality (Remainder Estimation)

The remainder after $n$ terms of the Taylor series satisfies:

|R_n(x)| \le \frac{M|x - a|^{n+1}}{(n+1)!}

Where $M$ is an upper bound for $|f^{(n+1)}(z)|$ for $z$ between $a$ and $x$ .

Example Use the Maclaurin series for $e^x$ to approximate $e^{0.1}$ with error less than $10^{-8}$ .

E^{0.1} = \sum_{n=0}^{N} \frac{(0.1)^n}{n!} + R_N

We need $|R_N| \le \frac{M \cdot (0.1)^{N+1}}{(N+1)!} \lt 10^{-8}$ .

Since $f^{(n+1)}(x) = e^x \le e^{0.1} \lt 1.2$ for $x \in [0, 0.1]$ :

\frac{1.2 \cdot (0.1)^{N+1}}{(N+1)!} \lt 10^{-8}

For $N = 3$ : $\displaystyle\frac{1.2 \cdot 10^{-4}}{24} = 5 \times 10^{-6}$ (too large).

For $N = 5$ : $\displaystyle\frac{1.2 \cdot 10^{-6}}{720} \approx 1.67 \times 10^{-9} \lt 10^{-8}$ .

So using 6 terms (up to $n = 5$ ):

E^{0.1} \approx 1 + 0.1 + \frac{0.01}{2} + \frac{0.001}{6} + \frac{10^{-4}}{24} + \frac{10^{-5}}{120} \approx 1.1051708

Deriving Maclaurin Series by Substitution

You do not need to compute derivatives from scratch every time. If you know the series for $e^u$ You can substitute $u = -x^2$ to get the series for $e^{-x^2}$ .

Example Find the Maclaurin series for $\cos(x^2)$ .

Substitute $u = x^2$ into the series for $\cos u$ :

\cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n (x^2)^{2n}}{(2n)!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n}}{(2n)!} = 1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \cdots

Operations on Power Series

Multiplication of Power Series

If $\sum a_n x^n$ and $\sum b_n x^n$ both have radius $R$ Then their Cauchy product also has radius $R$ :

\left(\sum_{n=0}^{\infty} a_n x^n\right)\!\left(\sum_{n=0}^{\infty} b_n x^n\right) = \sum_{n=0}^{\infty} c_n x^n

Where $c_n = \sum_{k=0}^{n} a_k b_{n-k}$ .

Example Find the Maclaurin series for $\ln(1 + x)$ by integrating the geometric series.

\frac{1}{1 - (-x)} = \sum_{n=0}^{\infty} (-x)^n = \sum_{n=0}^{\infty} (-1)^n x^n, \quad |x| \lt 1

Integrate term by term:

\int_0^x \frac{1}{1 + t}\, dt = \sum_{n=0}^{\infty} (-1)^n \int_0^x t^n\, dt = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1}

\ln(1 + x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n}, \quad |x| \lt 1

Common Pitfalls

Confusing sequences and series. A sequence is a list; a series is a sum. A convergent sequence does not imply a convergent series (e.g., $a_n = \frac{1}{n}$ converges to 0, but $\sum \frac{1}{n}$ diverges).
Using the $n$ Th-term test incorrectly. $\lim a_n = 0$ does not prove convergence (e.g., harmonic series). The test only detects divergence.
Forgetting to check endpoints of the interval of convergence for power series. The ratio test always gives $L = 1$ at the endpoints, so you must use a different test.
Misidentifying the center of a Taylor series. For $\sum c_n(x - 3)^n$ The center is $a = 3$ .
Applying the ratio test when $L = 1$ . The test is inconclusive; use a different test (comparison, integral, alternating series).
Confusing absolute and conditional convergence. An alternating harmonic series converges conditionally, not absolutely. Only absolutely convergent series can be freely rearranged.
Computing Taylor series coefficients incorrectly. Always use $c_n = \frac{f^{(n)}(a)}{n!}$ not just $f^{(n)}(a)$ . Forgetting to divide by $n!$ is a common mistake.
Assuming convergence at endpoints. The interval of convergence may be open, closed, or half-open at each endpoint. You must test each one individually.

Practice Questions

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{2^n}{n!}$ converges or diverges.
Find the interval of convergence for $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^n}{3^n}$ .
Find the Maclaurin series for $f(x) = x e^x$ .
How many terms of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ are needed to approximate the sum with error less than $0.01$ ?
Use the limit comparison test to determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{5}{2n^2 - 3n + 1}$ converges.
Find the Taylor series for $\ln x$ centered at $a = 1$ and determine its radius of convergence.
Find the Maclaurin series for $\arctan x$ by integrating the geometric series.
Use the Maclaurin series for $\cos x$ to approximate $\cos(0.2)$ with error less than $10^{-6}$ .
Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{n}{e^n}$ converges using the ratio test.
Find the Maclaurin series for $\frac{x}{1-x^2}$ and determine its interval of convergence.
Express $0.\overline{271}$ as a fraction using geometric series.
Use the alternating series estimation theorem to bound the error in approximating $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}$ by its first 10 terms.

Practice Problems

Question 1: Taylor series expansion

Find the first four nonzero terms of the Maclaurin series for $f(x) = \ln(1 + x)$ and use it to approximate $\ln(1.1)$ .

Answer

$f(0) = \ln(1) = 0$ .

$f'(x) = \frac{1}{1+x}$ , $f'(0) = 1$ .

$f''(x) = \frac{-1}{(1+x)^2}$ , $f''(0) = -1$ .

$f'''(x) = \frac{2}{(1+x)^3}$ , $f'''(0) = 2$ .

$f^{(4)}(x) = \frac{-6}{(1+x)^4}$ , $f^{(4)}(0) = -6$ .

Maclaurin series: $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$

For $\ln(1.1)$ with $x = 0.1$ : $\ln(1.1) \approx 0.1 - 0.005 + 0.000333 - 0.000025 = 0.095308$ .

Actual: $\ln(1.1) \approx 0.09531$ . The approximation is accurate to 5 decimal places.

Question 2: Ratio test

Determine the radius of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{(2x)^n}{n!}$ .

Answer

$a_n = \frac{2^n x^n}{n!}$ .

$\displaystyle L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{2^{n+1} x^{n+1} / (n+1)!}{2^n x^n / n!}\right| = \lim_{n \to \infty} \frac{2|x|}{n+1} = 0$ .

Since $L = 0 \lt 1$ for all $x$ The radius of convergence is $R = \infty$ . The series converges for all real $x$ . (This is the Maclaurin series for $e^{2x}$ .)

Question 3: Alternating series

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}$ converges absolutely, converges conditionally, or diverges.

Answer

Alternating series test: $a_n = \frac{1}{\sqrt{n}}$ is positive, decreasing, and $\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$ . The alternating series converges.

Absolute convergence: $\sum \frac{1}{\sqrt{n}}$ is a p-series with $p = 1/2 \lt 1$ So it diverges.

Therefore, the series converges conditionally (but not absolutely).

Question 4: Power series representation

Find the power series representation for $\displaystyle\frac{1}{(1-x)^2}$ and determine its interval of convergence.

Answer

We know $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ for $|x| \lt 1$ .

Differentiate both sides: $\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} nx^{n-1} = \sum_{n=0}^{\infty} (n+1)x^n$ .

Interval of convergence: $|x| \lt 1$ Or $(-1, 1)$ .

Check endpoints: at $x = 1$ Series is $\sum (n+1)$ which diverges. At $x = -1$ Series is $\sum (-1)^n(n+1)$ which diverges by the divergence test.

Question 5: Telescoping series

Evaluate $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ .

Answer

Partial fraction decomposition: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ .

$S_N = \sum_{n=1}^{N} \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \cdots - \frac{1}{N+1}$ .

All intermediate terms cancel (telescoping): $S_N = 1 - \frac{1}{N+1}$ .

$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \lim_{N \to \infty} \left(1 - \frac{1}{N+1}\right) = 1$ .

flowchart TD
    A[5_Sequences And Series] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

This topic covers the mathematical techniques and concepts related to sequences and series, including key theorems, methods, and problem-solving approaches.

Key concepts include:

arithmetic and geometric sequences
series and sigma notation
recurrence relations
convergence tests
mathematical induction

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Intuition

The infinite jigsaw puzzle: A series is like trying to complete an infinite jigsaw puzzle. Each piece (term) adds to the picture, and convergence means the puzzle eventually looks complete from a distance.

Why it matters: Series allow us to represent complex functions as simple infinite sums, making calculations easier and enabling approximations in physics, engineering, and computer science.

The key insight: An infinite series converges when the terms shrink fast enough that their sum approaches a finite value.

Cross-References

Limits and Continuity — Convergence of sequences and series is defined using the same limit concepts as continuity.
Derivatives — Taylor and Maclaurin series express functions as infinite sums of derivative-based terms.
Integrals — Term-by-term integration of power series connects series convergence to integral calculus.
AP Physics — Work, Energy, and Power: Series expansions approximate work and energy calculations when closed-form integrals are unavailable.
AP Physics — Kinematics: Taylor series approximate position and velocity functions, enabling linearisation of complex motions near a point.

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Sequences and Series | AP - Wyatt's Notes

Sequences (CED BC Unit 10)

Bounded and Monotone Sequences

Common Sequences

Proof: n!nn→0\frac{n!}{n^n} \to 0nnn!​→0

Proof: ln⁡nn→0\frac{\ln n}{n} \to 0nlnn​→0

Series (CED BC Unit 10)

Partial Sums

The nnnTh-Term Test (Divergence Test)

Common Pitfalls

Practice Questions

Practice Problems

Summary

Worked Examples

Intuition

Cross-References

Proof: $\frac{n!}{n^n} \to 0$

Proof: $\frac{\ln n}{n} \to 0$

The $n$ Th-Term Test (Divergence Test)