Sequences and Series

A sequence is an ordered list of numbers $a_1, a_2, a_3, \ldots$ written $\{a_n\}_{n=1}^{\infty}$ Or $\{a_n\}$. Each $a_n$ is called a term of the sequence. A series is the sum of The terms of a sequence: $\displaystyle\sum_{n=1}^{\infty} a_n$.

A sequence converges to a limit $L$ if $\displaystyle\lim_{n \to \infty} a_n = L$. Otherwise it diverges.

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An arithmetic sequence has a constant common difference $d$ between consecutive terms:

$$a_n = a_1 + (n - 1)d$$

Where $a_1$ is the first term and $d = a_{n+1} - a_n$ for all $n$.

The sum of the first $n$ terms is:

$$S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}\bigl[2a_1 + (n-1)d\bigr]$$

Proof. Write the sum forward and backward:

$$S_n = a_1 + (a_1 + d) + (a_1 + 2d) + \cdots + a_n$$ $$S_n = a_n + (a_n - d) + (a_n - 2d) + \cdots + a_1$$

Adding: $2S_n = n(a_1 + a_n)$ Hence $S_n = \dfrac{n}{2}(a_1 + a_n)$.

Example. Find the sum of the first 50 positive odd numbers.

$a_1 = 1$, $d = 2$, $a_{50} = 1 + 49 \cdot 2 = 99$.

$$S_{50} = \frac{50}{2}(1 + 99) = 2500$$

The arithmetic mean of $a$ and $b$ is $\dfrac{a + b}{2}$. In an arithmetic sequence, each term is the arithmetic mean of its neighbours.

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A geometric sequence has a constant common ratio $r$ between consecutive terms:

$$a_n = a_1 \cdot r^{n-1}$$

Where $a_1$ is the first term and $r = \dfrac{a_{n+1}}{a_n}$ for all $n$.

For $r \ne 1$:

$$S_n = \frac{a_1(1 - r^n)}{1 - r} = \frac{a_1(r^n - 1)}{r - 1}$$

Proof. $S_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1}$.

$$rS_n = a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1} + a_1 r^n$$

Subtracting: $S_n - rS_n = a_1 - a_1 r^n$.

$$S_n(1 - r) = a_1(1 - r^n) \implies S_n = \frac{a_1(1 - r^n)}{1 - r}$$

If $|r| \lt 1$ The infinite geometric series converges:

$$S_{\infty} = \sum_{n=1}^{\infty} a_1 r^{n-1} = \frac{a_1}{1 - r}$$

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

Example. Express $0.\overline{7}$ as a fraction.

$$0.\overline{7} = \frac{7}{10} + \frac{7}{100} + \frac{7}{1000} + \cdots$$

This is a geometric series with $a_1 = \dfrac{7}{10}$ and $r = \dfrac{1}{10}$.

$$S_{\infty} = \frac{7/10}{1 - 1/10} = \frac{7/10}{9/10} = \frac{7}{9}$$

Example. Evaluate $\displaystyle\sum_{n=0}^{\infty} \frac{3}{4^n}$.

$a_1 = 3$, $r = \dfrac{1}{4}$.

$$S_{\infty} = \frac{3}{1 - 1/4} = \frac{3}{3/4} = 4$$

The geometric mean of positive numbers $a$ and $b$ is $\sqrt{ab}$. In a geometric sequence with positive terms, each term is the geometric mean of its neighbours.

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$$\sum_{k=m}^{n} a_k = a_m + a_{m+1} + \cdots + a_n$$$$\sum_{k=m}^{n} c \cdot a_k = c\sum_{k=m}^{n} a_k$$ $$\sum_{k=m}^{n}(a_k + b_k) = \sum_{k=m}^{n} a_k + \sum_{k=m}^{n} b_k$$

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

Caution. If $\lim_{n \to \infty} a_n = 0$ The series may or may not converge. The harmonic series $\sum \dfrac{1}{n}$ diverges despite its terms tending to zero.

If $0 \le a_n \le b_n$ for all $n$:

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

For a series $\sum a_n$ with $a_n \ne 0$:

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

• $L \lt 1$: the series converges absolutely.
• $L \gt 1$: the series diverges.
• $L = 1$: the test is inconclusive.

Example. Does $\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{2^n}$ converge?

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

The series converges.

If $f$ is continuous, positive, and decreasing on $[1, \infty)$ Then $\sum f(n)$ converges if and only if $\displaystyle\int_1^{\infty} f(x)\,dx$ converges.

Example. The harmonic series diverges because:

$$\int_1^{\infty} \frac{1}{x}\,dx = \lim_{b \to \infty} \ln b = \infty$$

An alternating series $\sum (-1)^{n+1} a_n$ (or $\sum (-1)^n a_n$) converges if:

1. $a_n$ is decreasing: $a_{n+1} \le a_n$ for all $n$.
2. $\displaystyle\lim_{n \to \infty} a_n = 0$.

Example. $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converges (the alternating harmonic Series) since $\dfrac{1}{n}$ decreases to $0$.

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For $n \in \mathbb{Z}^+$:

$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

Where the binomial coefficient is:

$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$

For any $n \in \mathbb{R}$ and $|x| \lt 1$:

$$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$

This is an infinite series that converges for $|x| \lt 1$ (and at the endpoints depending on $n$).

Example. Expand $(1 + x)^{-2}$ up to the term in $x^3$.

$$(1 + x)^{-2} = 1 + (-2)x + \frac{(-2)(-3)}{2}x^2 + \frac{(-2)(-3)(-4)}{6}x^3 + \cdots$$ $$= 1 - 2x + 3x^2 - 4x^3 + \cdots$$

Example. Find the coefficient of $x^3$ in the expansion of $(2 - 3x)^{1/2}$.

$$(2 - 3x)^{1/2} = \sqrt{2}\left(1 - \frac{3x}{2}\right)^{1/2}$$ $$= \sqrt{2}\left[1 + \frac{1}{2}\!\left(-\frac{3x}{2}\right) + \frac{(1/2)(-1/2)}{2}\!\left(-\frac{3x}{2}\right)^{\!2} + \frac{(1/2)(-1/2)(-3/2)}{6}\!\left(-\frac{3x}{2}\right)^{\!3} + \cdots\right]$$

The $x^3$ coefficient:

$$\sqrt{2} \cdot \frac{-3/8}{6} \cdot \left(-\frac{27}{8}\right) = \sqrt{2} \cdot \frac{81}{384} = \frac{27\sqrt{2}}{128}$$

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The Maclaurin series of a function $f$ is its Taylor series expansion about $x = 0$:

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\,x^n = f(0) + f"(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$

Example. Find the Maclaurin series of $e^x$.

$$f(0) = 1, \quad f'(0) = 1, \quad f''(0) = 1, \quad \ldots, \quad f^{(n)}(0) = 1$$ $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

Example. Find the Maclaurin series of $\sin x$.

$$f(0) = 0, \quad f'(0) = 1, \quad f''(0) = 0, \quad f'''(0) = -1, \quad f^{(4)}(0) = 0, \quad \ldots$$ $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$

Example. Find the Maclaurin series of $x^2 e^x$.

Since $e^x = \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}$ Multiplying by $x^2$:

$$x^2 e^x = \sum_{n=0}^{\infty} \frac{x^{n+2}}{n!} = x^2 + x^3 + \frac{x^4}{2!} + \frac{x^5}{3!} + \cdots$$

Example. Find the Maclaurin series of $\dfrac{1}{1 + x^2}$.

$$\frac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + \cdots = \sum_{n=0}^{\infty} (-1)^n x^{2n}$$

Integrating term by term gives the series for $\arctan x$:

$$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$

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The Taylor series of $f$ about $x = a$ is:

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

This reduces to the Maclaurin series when $a = 0$.

The $n$-th degree Taylor polynomial of $f$ about $a$ is:

$$T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^k$$

The error in approximating $f(x)$ by $T_n(x)$ is bounded by:

$$|R_n(x)| = \left|f(x) - T_n(x)\right| \le \frac{M|x - a|^{n+1}}{(n+1)!}$$

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

Example. Approximate $\sqrt{e}$ using a 3rd degree Maclaurin polynomial and bound the error.

$T_3(x) = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}$.

$T_3(0.5) = 1 + 0.5 + 0.125 + 0.020833\ldots = 1.64583\ldots$

True value: $e^{0.5} \approx 1.64872$.

For the error bound, $|f^{(4)}(t)| = e^t \le e^{0.5}$ for $0 \le t \le 0.5$:

$$|R_3(0.5)| \le \frac{e^{0.5} \cdot (0.5)^4}{24} \approx \frac{1.649 \cdot 0.0625}{24} \approx 0.0043$$

Actual error: $|1.64872 - 1.64583| \approx 0.0029$ Which is within the bound.

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Mathematical induction proves a statement $P(n)$ for all integers $n \ge n_0$:

1. Base case: Verify $P(n_0)$ is true.
2. Inductive hypothesis: Assume $P(k)$ is true for some arbitrary $k \ge n_0$.
3. Inductive step: Using the hypothesis, prove $P(k + 1)$ is true.
4. Conclusion: By the principle of mathematical induction, $P(n)$ is true for all $n \ge n_0$.

Example. Prove that $\displaystyle\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ for all $n \in \mathbb{Z}^+$.

Base case ($n = 1$): $\dfrac{1 \cdot 2 \cdot 3}{6} = 1$. True.

Inductive hypothesis: Assume $\displaystyle\sum_{k=1}^{j} k^2 = \frac{j(j+1)(2j+1)}{6}$ for some $j \ge 1$.

Inductive step:

$$\sum_{k=1}^{j+1} k^2 = \sum_{k=1}^{j} k^2 + (j+1)^2 = \frac{j(j+1)(2j+1)}{6} + (j+1)^2$$ $$= \frac{(j+1)}{6}\bigl[j(2j+1) + 6(j+1)\bigr] = \frac{(j+1)}{6}\bigl[2j^2 + 7j + 6\bigr]$$ $$= \frac{(j+1)(j+2)(2j+3)}{6}$$

This is the formula with $n = j + 1$. Hence $P(j+1)$ is true. By induction, the result holds for All $n \in \mathbb{Z}^+$.

Example. Prove that $3^n - 1$ is divisible by $2$ for all $n \in \mathbb{N}$.

Base case ($n = 0$): $3^0 - 1 = 0$ Which is divisible by $2$. True.

Inductive hypothesis: $3^k - 1 = 2m$ for some $m \in \mathbb{Z}$.

Inductive step:

$$3^{k+1} - 1 = 3 \cdot 3^k - 1 = 3(2m + 1) - 1 = 6m + 3 - 1 = 6m + 2 = 2(3m + 1)$$

This is divisible by $2$. By induction, the result holds.

Example. Prove that $2^n \gt n$ for all $n \in \mathbb{Z}^+$.

Base case ($n = 1$): $2^1 = 2 \gt 1$. True.

Inductive hypothesis: $2^k \gt k$ for some $k \ge 1$.

Inductive step:

$$2^{k+1} = 2 \cdot 2^k \gt 2k \ge k + 1$$

(The last inequality holds since $k \ge 1$.) Therefore $2^{k+1} \gt k + 1$. By induction, $2^n \gt n$ For all positive integers $n$.