Sequences and Series

Example

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

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

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

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 $+\infty$.

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:

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

Example

Evaluate $\displaystyle\sum_{n=1}^{\infty} \frac{3}{2^n}$.

Rewrite as $\displaystyle\sum_{n=1}^{\infty} 3\left(\frac{1}{2}\right)^n$. Here $a = \frac{3}{2}$ And $r = \frac{1}{2}$.

Every repeating decimal can be expressed as a rational number using geometric series.

Example

Express $0.\overline{37}$ as a fraction.

This is a geometric series with $a = \frac{37}{100}$ and $r = \frac{1}{100}$.

A telescoping series has terms that cancel in pairs when the partial sum is expanded.

Example

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

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

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: $\frac{1}{(n-1)(n+1)} = \frac{1}{2}\!\left(\frac{1}{n-1} - \frac{1}{n+1}\right)$.

Most terms cancel, leaving:

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

Either both converge or both diverge.

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

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.

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

• 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.

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.

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

• 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 $p = 2$):

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

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

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

For $\displaystyle\sum a_n$Compute:

• 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 $\displaystyle\sum_{n=1}^{\infty} \frac{n!}{10^n}$ converges.

Since $L = \infty \gt 1$The series diverges.

Example

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

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

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

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

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.

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:

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.

• 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.

A power series centered at $a$ is:

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?

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 $L = 1$).

Example

Find the interval of convergence for $\displaystyle\sum_{n=0}^{\infty} \frac{(x - 2)^n}{n!}$.

Apply the ratio test:

Since $L = 0 \lt 1$ for all $x$The series converges for all real numbers. The interval of Convergence is $(-\infty, \infty)$ with $R = \infty$.

Example

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

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: $[0, 2)$.

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

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

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

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

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 $f(x)$.

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

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 $10^{-8}$.

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 $x \in [0, 0.1]$:

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 $n = 5$):

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 $\cos u$:

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

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

Example

Find the Maclaurin series for $e^x \sin x$ up to the $x^5$ term.

Multiplying and collecting terms up to $x^5$:

Example

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

Integrate term by term: