Unit 10

Infinite Sequences and Series: the one-page sheet

10.1 Convergent and Divergent Series

A series ∑ a_n converges to S when its partial sums S_n = a_1 + ⋯ + a_n approach S ; otherwise it diverges. Find terms from partial sums by subtracting, keep the terms and the partial sums apart, and say what converges: the series.

An infinite series $\sum_{n=1}^{\infty} a_n$ converges to $S$ exactly when its partial sums $S_n$ approach $S$ . Two sequences are in play: the terms $a_n$ and the partial sums $S_n$ . Keep them apart.
From partial sums back to terms: $a_1 = S_1$ and $a_n = S_n - S_{n-1}$ for $n \ge 2$ .
Say \"the series converges\" or \"the series diverges\". The terms of a convergent series approach 0, not the sum.
Constants factor out of a convergent series; adding or removing finitely many terms changes the sum but never whether the series converges.

the partial sum S_n = a_1 + a_2 + ⋯ + a_n a new sequence, S_1, S_2, S_3, … the series converges and its sum is S the series diverges and has no sum

infinite series: A sum with infinitely many terms, $\sum_{n=1}^{\infty} a_n$ $= a_1$ $+ a_2$ $+ \cdots$ It converges to $S$ when its partial sums approach $S$ , and diverges when they have no finite limit.
partial sum: The sum of the first $n$ terms of a series, $S_n$ $= a_1 + a_2 + \cdots + a_n$ . The partial sums form a sequence whose limit, if it exists, is the sum of the series.

10.2 Geometric Series

A geometric series ∑ ar^n has a constant ratio r between terms. It converges exactly when |r| < 1 , to (first term)/(1 − r) , where the first term is the first one actually in the sum. If |r| ≥ 1 it diverges.

$\sum_{n=0}^{\infty} ar^n$ $= \frac{a}{1 - r}$ when $|r| < 1$ ; it diverges when $|r| \ge 1$ (and $a \ne 0$ ).
In the formula, $a$ is the first term of the sum as written: $\sum_{n=1}^{\infty} 5\left(\frac{1}{4}\right)^n$ has $a = \frac{5}{4}$ , not 5.
Find $r$ by dividing a term by the one before it; rewrite $\frac{2^n}{3^{n+1}}$ as $\frac{1}{3}\left(\frac{2}{3}\right)^n$ to see $a$ and $r$ .
With $x$ in the ratio, the series converges for the $x$ where $|r| < 1$ , and its sum is a function of $x$ .

S_n = a +ar +⋯ +ar^(n − 1) = (a(1 − r^n))/(1 − r) S_n → a/(1 − r) : the series converges the partial sums have no limit: the series diverges a is the first term actually in the sum, not the value at n = 0

geometric series: A series $\sum ar^n$ in which each term is the previous one times the common ratio $r$ . It converges exactly when $|r| < 1$ , to $\frac{a}{1 - r}$ , where $a$ is the first term in the sum; it diverges when $|r| \ge 1$ (for $a \ne 0$ ).

10.3 The nth Term Test for Divergence

If the terms of a series do not approach 0, the series diverges: that is the nth term test. If the terms do approach 0, the test is inconclusive and another test must decide. Say \"the series diverges\", give the limit of the terms, and show it is not 0.

The nth term test: if $\lim_{n \to \infty} a_n \ne 0$ (or the limit does not exist), then $\sum a_n$ diverges.
It is a test for divergence only. When $\lim a_n = 0$ it gives no conclusion: $\sum \frac{1}{n^2}$ converges and $\sum \frac{1}{n}$ diverges, and both have terms approaching 0.
Write it fully: the limit of the terms, its value, \" $\ne 0$ \", the test's name, and \"the series diverges\".

S_n → S and S_(n − 1) → S , so a_n = S_n − S_(n − 1) → 0 the series cannot converge: it diverges the test says nothing; use another test

nth term test: If $\lim_{n \to \infty} a_n \ne 0$ or the limit does not exist, then $\sum a_n$ diverges. A test for divergence only: when $\lim a_n = 0$ it gives no conclusion.

10.4 The Integral Test

For a series of positive terms a_n = f(n) with f positive, continuous and decreasing, compute lim as b → ∞, ∫ from k to b of f(x) dx . A finite limit means the series converges; an infinite one means it diverges. Check the conditions, keep limit notation, and never report the integral as the sum.

Integral test: if $f$ is positive, continuous and decreasing on $[k, \infty)$ and $a_n = f(n)$ , then $\sum_{n=k}^{\infty} a_n$ and $\int_k^{\infty} f(x)\,dx$ both converge or both diverge.
State the three conditions every time. \"Decreasing\" usually needs $f'(x) < 0$ ; it may hold only from some $x$ on, and then the test starts there.
The integral's value is not the series' sum. The test decides convergence only.
Write the improper integral as a limit, $\lim_{b \to \infty} \int_k^{b} f(x)\,dx$ , to the end.

each term is the area of a box of width 1 that fits beside the curve a_2 +a_3 +⋯ ≤ ∫ from 1 to ∞ of f(x) dx : a finite integral holds the series down a_1 +⋯ +a_n ≥ ∫ from 1 to n + 1 of f(x) dx : an infinite integral pushes the series up both converge or both diverge

integral test: If $f$ is positive, continuous and decreasing on $[k, \infty)$ and $a_n = f(n)$ , then $\sum_{n=k}^{\infty} a_n$ and $\int_k^{\infty} f(x)\,dx$ both converge or both diverge. The integral's value is not the series' sum.

10.5 Harmonic Series and p-Series

A p -series ∑ 1/(n^p) converges exactly when p > 1 ; the integral test proves it. The harmonic series ( p = 1 ) diverges and the alternating harmonic series converges. Simplify a series to the form 1/(n^p) , then compare p with 1.

$\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \le 1$ .
The harmonic series $\sum \frac{1}{n}$ ( $p = 1$ ) diverges, though its terms approach 0. The alternating harmonic series $\sum \frac{(-1)^{n+1}}{n}$ converges.
Rewrite before deciding: $\frac{1}{\sqrt[3]{n}} = \frac{1}{n^{1/3}}$ , $\frac{\sqrt{n}}{n^2} = \frac{1}{n^{3/2}}$ , $\frac{n^2}{n^3} = \frac{1}{n}$ . Name the series and its $p$ when you use it.

f is positive, continuous and decreasing on [1, ∞) the p -series converges the p -series diverges the series diverges by the nth term test

p-series: A series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ . It converges if $p > 1$ and diverges if $p \le 1$ , which the integral test proves.
harmonic series: The $p$ -series with $p = 1$ , $\sum_{n=1}^{\infty} \frac{1}{n}$ , which diverges although its terms approach 0. The alternating harmonic series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converges.

10.6 Comparison Tests

For a series with positive terms, compare it with a known geometric series or p -series. Direct comparison: smaller than a convergent series converges, larger than a divergent series diverges. Limit comparison: if lim as n → ∞, (a_n)/(b_n) is finite and positive, both series do the same thing.

Compare positive terms with a series whose behavior is known and named: geometric with its ratio, or a $p$ -series with its $p$ .
Direct comparison needs the inequality in the useful direction: smaller than convergent, or larger than divergent. Larger than convergent, or smaller than divergent, proves nothing.
Choose $b_n$ by keeping the highest powers on top and bottom; use limit comparison when the inequality points the wrong way.
Write the limit in $n$ : $\lim_{n \to \infty} \frac{a_n}{b_n}$ , and say it is finite and positive.

its partial sums only increase, so they converge exactly when they stay below some ceiling each partial sum of ∑ a_n is below the sum of ∑ b_n , so ∑ a_n converges (direct comparison) the partial sums of ∑ a_n grow past every bound, so ∑ a_n diverges for large n , a_n is about L b_n , so the two series converge together or diverge together (limit comparison)

direct comparison test: For series with $0 \le a_n \le b_n$ : if $\sum b_n$ converges, so does $\sum a_n$ ; if $\sum a_n$ diverges, so does $\sum b_n$ .
limit comparison test: For series with positive terms: if $\lim_{n \to \infty} \frac{a_n}{b_n}$ is a finite positive number, then $\sum a_n$ and $\sum b_n$ both converge or both diverge.

10.7 The Alternating Series Test

A series whose terms alternate in sign converges when the sizes of its terms decrease and approach 0. Its partial sums zigzag around the sum with shrinking swings. If the sizes do not approach 0, the nth term test shows divergence; the alternating series test itself never shows divergence.

The alternating series test: if $a_n > 0$ , the $a_n$ decrease (from some point on), and $\lim_{n \to \infty} a_n = 0$ , then $\sum (-1)^n a_n$ converges.
Check and state all three conditions: alternating signs, decreasing sizes, a limit of 0.
The test only proves convergence. If the sizes do not approach 0, use the nth term test for divergence instead.
When the conditions hold from the first term on, the sum lies between any two consecutive partial sums.

each partial sum steps up, then down, then up again, so the partial sums zigzag each step is shorter than the last, so the zigzag never passes back beyond the previous turn the steps shrink to nothing, the zigzag closes in on one number, and the series converges

alternating series: A series whose terms alternate in sign, $\sum (-1)^n a_n$ with $a_n > 0$ . The alternating series test: if the $a_n$ decrease and approach 0, the series converges.

10.8 The Ratio Test

The ratio test compares a series with a geometric one. Find L = lim as n → ∞, |(a_(n + 1))/(a_n)| : if L < 1 the series converges, if L > 1 or L is infinite it diverges, and if L = 1 the test gives no conclusion. It works best on factorials and powers like 3^n .

Ratio test: $L$ $= \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$ . $L < 1$ : converges. $L > 1$ or infinite: diverges. $L = 1$ : no conclusion.
Put $a_{n+1}$ on top and simplify before taking the limit: $\frac{(n+1)!}{n!} = n + 1$ , $\frac{3^{n+1}}{3^n} = 3$ .
Use it for factorials and exponentials. For $p$ -series and ratios of polynomials the limit is 1, so compare instead.
Only the limit matters; early ratios above or below 1 change nothing.

far out, each term is about L times the one before, as in a geometric series with ratio L the terms shrink at least as fast as a convergent geometric series, so the series converges the terms grow, do not approach 0, and the series diverges the test decides nothing: use another test

ratio test: For a series $\sum a_n$ with nonzero terms, let $L$ $= \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$ . If $L < 1$ the series converges; if $L > 1$ or the limit is infinite it diverges; if $L = 1$ the test gives no conclusion.

10.9 Absolute and Conditional Convergence

Every series is absolutely convergent ( ∑ |a_n| converges), conditionally convergent ( ∑ a_n converges but ∑ |a_n| diverges), or divergent. Absolute convergence implies convergence, and an absolutely convergent series keeps its sum when its terms are rearranged.

Absolutely convergent: $\sum |a_n|$ converges. Conditionally convergent: $\sum a_n$ converges but $\sum |a_n|$ diverges.
Absolute convergence implies convergence.
To classify: test $\sum |a_n|$ first. If it converges, you are done (absolute). If not, test $\sum a_n$ itself, usually with the alternating series test.
A divergent $\sum |a_n|$ alone decides nothing about $\sum a_n$ .

∑ |a_n| is a series of positive terms, which the comparison, integral and ratio tests can decide ∑ a_n converges too (absolutely): signs can only cancel, never add the convergence comes from the signs cancelling: conditional convergence its terms can be rearranged in any order without changing the sum

absolute convergence: A series $\sum a_n$ converges absolutely when $\sum |a_n|$ converges. Absolute convergence implies convergence, and the sum does not change when the terms are rearranged.
conditional convergence: A series converges conditionally when $\sum a_n$ converges but $\sum |a_n|$ diverges, as $\sum \frac{(-1)^{n+1}}{n}$ does.

10.10 The Alternating Series Error Bound

For a series that alternates with sizes decreasing to 0, the sum lies within a_(n + 1) , the first omitted term, of S_n : |S − S_n| ≤ a_(n + 1) . State the conditions, compute the bound, and compare it with the required accuracy using " ≤ … < ".

If a series alternates and its sizes decrease to 0, then $|S - S_n| \le a_{n+1}$ : the error is at most the first term left out.
State the conditions every time; the bound means nothing without them.
Write the result as an inequality: "error $\le \frac{1}{720} < 0.002$ ". The error is at most the bound, not equal to it.
The sign of the first omitted term says whether $S_n$ is too big or too small.

the sum always lies between two consecutive partial sums, S_n and S_(n + 1) the distance from S_n to the sum is at most a_(n + 1) : |S − S_n| ≤ a_(n + 1) tells the direction: if it is negative, S_n overestimates the sum; if positive, it underestimates

error bound: A number that the size of an error cannot exceed: if $E$ is an error bound for an approximation, then $|\text{actual} - \text{approximation}| \le E$ .
alternating series error bound: For an alternating series whose sizes $a_n$ decrease to 0, the error of the partial sum $S_n$ is at most the first omitted term: $|S - S_n| \le a_{n+1}$ .

10.11 Taylor Polynomials

The n th-degree Taylor polynomial for f about x = a is ∑ for k = 0 to n of (f^((k))(a))/(k!) (x − a)^k . It matches f and its first n derivatives at a , so it approximates f near a . About a = 0 it is called a Maclaurin polynomial.

$P_n(x)$ $= f(a)$ $+ f'(a)(x - a)$ $+ \frac{f''(a)}{2!}(x - a)^2$ $+ \cdots$ $+ \frac{f^{(n)}(a)}{n!}(x - a)^n$ .
The center is where the derivatives are taken: powers of $(x - a)$ , derivatives at $a$ .
$P_1$ is the tangent line; each higher term matches one more derivative at the center.
A Taylor polynomial ends at its last term: write it out with no " $+ \cdots$ ".

it fits f well very near a but misses the bending the polynomial's second derivative at a is f″(a) , so it bends the same way f does the polynomial matches more derivatives at a and usually stays close to f over a wider stretch

Taylor polynomial: The polynomial $P_n(x)$ $= \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^k$ , which matches $f$ and its first $n$ derivatives at the center $x = a$ . A Maclaurin polynomial is a Taylor polynomial about $x = 0$ .

10.12 The Lagrange Error Bound

The Lagrange error bound caps the error of a Taylor polynomial: |f(x) − P_n(x)| ≤ M/((n + 1)!) |x − a|^(n + 1) , where M is the largest value of |f^((n + 1))| between a and x . Justify M , compute the bound, and compare it with the accuracy required.

$\left|f(x) - P_n(x)\right| \le \frac{\max\left|f^{(n+1)}(z)\right|}{(n+1)!}|x - a|^{n+1}$ , the maximum taken for $z$ between $a$ and $x$ .
Say where the maximum is and why: " $e^z$ is increasing, so on $[0, 0.5]$ its maximum is $e^{0.5} < 2$ ."
The bound is a ceiling: write "error $\le$ ", never "error $=$ ".
If the values form an alternating sum with sizes decreasing to 0, the alternating series error bound may be quicker.

the difference f − P_n grows only through the derivative after those, f^((n + 1)) the error can be no bigger than M/((n + 1)!) |x − a|^(n + 1) "error ≤ bound < given number" shows the estimate is good enough

Lagrange error bound: A bound on the error of a Taylor polynomial: $\left|f(x) - P_n(x)\right| \le \frac{M}{(n+1)!}|x - a|^{n+1}$ , where $M$ is the largest value of $\left|f^{(n+1)}(z)\right|$ for $z$ between the center $a$ and $x$ .

10.13 Radius and Interval of Convergence

For a power series centered at a , the ratio test gives |x − a| < R : the radius R and an open interval. The ratio test limit is 1 at both endpoints, so test each one separately and include it only if the series converges there.

A power series $\sum a_n (x - a)^n$ converges only at its center, for every real $x$ , or on an interval of radius $R$ around the center.
The ratio test gives the radius; it never decides the endpoints.
Test each endpoint separately: substitute, simplify, and use the $p$ -series, alternating series or nth term test.
Radius and interval are different answers: $R = 4$ , but the interval is $[-3, 5)$ .

the limit is a multiple of |x − a| the radius R and the open interval (a − R, a + R) , where the series converges absolutely substitute each endpoint, name the series you get, and test it with another test write the interval with a bracket for an included endpoint and a parenthesis for one left out

power series: A series of the form $\sum_{n=0}^{\infty} a_n (x - a)^n$ , centered at $x = a$ . For each $x$ it is a series of numbers; its sum is a function of $x$ wherever it converges.
radius of convergence: The number $R$ such that a power series centered at $a$ converges (absolutely) when $|x - a| < R$ and diverges when $|x - a| > R$ . It is 0 if the series converges only at its center and infinite if it converges for every $x$ .
interval of convergence: The set of all $x$ where a power series converges: the open interval $(a - R, a + R)$ from the radius, plus each endpoint at which the series, tested on its own, converges.

10.14 Taylor and Maclaurin Series

The Taylor series of f about a is ∑ (f^((n))(a))/(n!) (x − a)^n ; its partial sums are the Taylor polynomials. The Maclaurin series of e^x , sin x and cos x converge to the function for every x , and 1/(1 − x) = ∑ x^n for |x| < 1 .

Taylor series of $f$ about $a$ : $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n$ . About $a = 0$ it is a Maclaurin series.
$e^x = \sum \frac{x^n}{n!}$ , $\sin x$ $= \sum \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ , $\cos x$ $= \sum \frac{(-1)^n x^{2n}}{(2n)!}$ , for every $x$ .
$\frac{1}{1 - x}$ $= \sum x^n$ $= 1 + x + x^2 + \cdots$ only for $-1 < x < 1$ .
A series is written with " $+ \cdots$ " or with $\sum$ and a general term; a Taylor polynomial is not.

the result is a power series, the Taylor series of f about a , and P_n is its partial sum the series equals the function on its interval of convergence many sums and limits can be read off, and other series can be built from them

Taylor series: The power series $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n$ of a function $f$ about $x = a$ ; its partial sums are the Taylor polynomials. A Maclaurin series is a Taylor series about $x = 0$ .

10.15 Representing Functions as Power Series

New power series come from known ones by substitution, multiplication by powers of x , and term-by-term differentiation or integration. These operations keep the radius of convergence, but the endpoints must be tested again. This is how the series for ln (1 + x) and arctan x are derived.

Substitution: replace $x$ by an expression in a known series, and the condition with it: $\frac{1}{1 + x^2} = \sum (-x^2)^n$ for $x^2 < 1$ .
Term by term: $\frac{d}{dx}\sum c_n x^n$ $= \sum n c_n x^{n-1}$ and $\int_0^x \sum c_n t^n\,dt$ $= \sum \frac{c_n x^{n+1}}{n + 1}$ .
The radius stays the same; endpoints may change, so test them again.
Differentiate with respect to $x$ , never $n$ ; when integrating, fix the constant with a known value such as $\ln 1 = 0$ .

its sum and its interval are known a series for a new function, valid where the substituted value meets the old condition a series for the derivative or integral with the same radius; test the endpoints again

term-by-term differentiation: Differentiating or integrating a power series one term at a time: $\frac{d}{dx}\sum c_n x^n$ $= \sum n c_n x^{n-1}$ and $\int_0^x \sum c_n t^n\,dt$ $= \sum \frac{c_n x^{n+1}}{n + 1}$ . The radius of convergence stays the same; the endpoints must be tested again.