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Series and Taylor polynomials

A series is a sum of infinitely many terms. Some have a finite value (they converge) and others grow without limit (they diverge), even though their terms keep getting smaller. This topic explains geometric and p-series, the convergence tests and the Taylor polynomial, which approximates a function with a polynomial: that is how calculators compute sines and exponentials.

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Series and partial sums

The series $\sum_{n} a_{n}$ converges to $S$ if its partial sums $S_{N} = a_{1} + \dots + a_{N}$ tend to $S$ . If $a_{n}$ does not tend to 0, the series surely diverges (term test); but $a_{n} \to 0$ is not enough: it is necessary, not sufficient.

Geometric: each term is the previous one times a fixed ratio $r$ . It converges if and only if $∣ r ∣ < 1$ :

n = 0 \sum \infty a r^{n} = \frac{a}{1 - r} (∣ r ∣ < 1)

1 + \frac{1}{2} + \frac{1}{4} + \dots = 2

: the partial sums approach 2 without passing it.

p-series: $\sum \frac{1}{n ^{p}}$ converges if and only if $p > 1$ . The harmonic series ( $p = 1$ ) diverges, though very slowly: about 12,000 terms are needed for the sum to pass 10.

Convergence tests

Test	What it says	When to use it
Ratio	$L = lim \frac{a _{n + 1}}{a _{n}}$ : converges if $L < 1$ , diverges if $L > 1$ ; $L = 1$ does not decide	factorials, powers $c^{n}$
Root	$L = lim n ∣ a_{n} ∣$ , same rules	terms of the form $(\dots)^{n}$
Comparison	if $0 \leq a_{n} \leq b_{n}$ and $\sum b_{n}$ converges, so does $\sum a_{n}$ (and conversely for divergence)	quotients of polynomials: compare with a p-series
Integral	$\sum f (n)$ and $\int^{\infty} f (x) d x$ converge or diverge together ( $f$ positive and decreasing)	functions easy to integrate
Leibniz	an alternating $\sum (- 1)^{n} b_{n}$ with $b_{n}$ decreasing and $b_{n} \to 0$ converges	alternating signs

A series converges absolutely if $\sum ∣ a_{n} ∣$ converges, and conditionally if it converges but $\sum ∣ a_{n} ∣$ does not.

Taylor polynomial

Near a point $x_{0}$ , a differentiable function is approximated by the polynomial that has the same derivatives at $x_{0}$ :

f (x) \approx P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( x _{0} )}{k !} (x - x_{0})^{k}

With $x_{0} = 0$ it is called Maclaurin. The most used ones:

e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots sin x = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots

Taylor polynomials of

sin x

at 0: the higher the order, the wider the interval where they match the curve.

The error is of the order of the first omitted term, $(x - x_{0})^{n + 1}$ : it grows away from $x_{0}$ . That is where the engineering approximations $sin x \approx x$ and $cos x \approx 1 - x^{2} /2$ for small angles (in radians) come from.

Common mistakes

Concluding that a series converges because $a_{n} \to 0$ (the harmonic series is the counterexample).
Applying the ratio test when it gives $L = 1$ and deciding something: with $L = 1$ another test is needed.
Using the geometric formula with $∣ r ∣ \geq 1$ , or with the wrong first term.
Forgetting the $k!$ in the denominator in Taylor, or using degrees instead of radians.

Worked examples

Example 1Geometric series

Does

\sum_{n = 0}^{\infty} (\frac{2}{3})^{n}

converge? If so, what is its value?

Solution

It is geometric with ratio $r = \frac{2}{3}$ and first term $a = 1$ . Since $∣ r ∣ < 1$ , it converges.
$\sum = \frac{1}{1 - 2/3} = 3$ .

It converges; sum 3.

Solve it in Simuveliq

In Math → Symbolic calculator, topic Series and differential equations, operation Series convergence: type (2/3)^n with From n = 0 and press Compute. The Answer is «The series converges (geometric)», and the steps give the sum, 3.

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Example 2The harmonic series

Does

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

converge?

Solution

$\frac{1}{n} \to 0$ : the term test does not decide. The ratio test gives $L = 1$ : neither.
It is a p-series with $p = 1$ , which is not greater than 1: it diverges. (With the integral: $\int_{1}^{\infty} \frac{d x}{x} = lim ln b = \infty$ .)

It diverges.

Solve it in Simuveliq

In Math → Symbolic calculator, topic Series and differential equations, operation Series convergence: type 1/n with From n = 1 and press Compute. The Answer is «The series diverges (p-series)».

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Example 3A convergent p-series

Does

\sum_{n = 1}^{\infty} \frac{1}{n ^{2}}

converge?

Solution

It is a p-series with $p = 2 > 1$ : it converges. (Its sum is $π^{2} /6 \approx 1.645$ , a classic result that does not follow from these tests.)

It converges.

Solve it in Simuveliq

In Math → Symbolic calculator, topic Series and differential equations, operation Series convergence: type 1/n^2 with From n = 1 and press Compute. The Answer is «The series converges (Raabe's test)»: since the ratio gives $L = 1$ , the calculator uses Raabe's test, another route to the same conclusion.

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Example 4Conditional convergence

Study

\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n}}{n}

Solution

It is alternating with $b_{n} = \frac{1}{n}$ , decreasing and with limit 0: it converges by Leibniz.
In absolute value it is the harmonic series, which diverges: the convergence is conditional (its sum is $- ln 2$ ).

It converges conditionally.

Solve it in Simuveliq

In Math → Symbolic calculator, topic Series and differential equations, operation Series convergence: type (-1)^n/n with From n = 1 and press Compute. The Answer is «The series converges (conditionally)».

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Example 5Ratio test

Does

\sum_{n = 1}^{\infty} \frac{n}{2 ^{n}}

converge?

Solution

$\frac{a _{n + 1}}{a _{n}} = \frac{n + 1}{2 ^{n + 1}} \cdot \frac{2 ^{n}}{n} = \frac{n + 1}{2 n} \to \frac{1}{2}$ .
$L = \frac{1}{2} < 1$ : it converges (absolutely). The power $2^{n}$ beats $n$ .

It converges absolutely.

Solve it in Simuveliq

In Math → Symbolic calculator, topic Series and differential equations, operation Series convergence: type n/2^n with From n = 1 and press Compute. The Answer is «The series converges (absolutely, by the ratio test)».

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Example 6Taylor polynomial of sine

Find the order-7 Maclaurin polynomial of

sin x

and use it to estimate

sin 0.5

Solution

Derivatives at 0: $sin 0 = 0$ , $cos 0 = 1$ , $- sin 0 = 0$ , $- cos 0 = - 1$ , and they repeat. Only odd powers remain:
$P_{7} (x) = x - \frac{x ^{3}}{6} + \frac{x ^{5}}{120} - \frac{x ^{7}}{5040}$ .
$P_{7} (0.5) = 0.5 - 0.0208333 + 0.0002604 - 0.0000016 = 0.4794255$ ; $sin 0.5 = 0.4794255$ : they agree to 7 decimals.

sin x \approx x - \frac{x ^{3}}{6} + \frac{x ^{5}}{120} - \frac{x ^{7}}{5040}

Solve it in Simuveliq

In the topic Calculus, operation Taylor series: type sin(x), Center = 0 and Order = 7; press Compute. The Answer is $- \frac{x ^{7}}{5040} + \frac{x ^{5}}{120} - \frac{x ^{3}}{6} + x$ ; the small graph draws $f$ and $P_{7}$ .

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Example 7Taylor away from the origin

Find the order-4 Taylor polynomial of

ln x

centered at

x_{0} = 1

Solution

$f (1) = 0$ , $f^{'} = \frac{1}{x} \to 1$ , $f^{''} = - \frac{1}{x ^{2}} \to - 1$ , $f^{'''} = \frac{2}{x ^{3}} \to 2$ , $f^{(4)} = - \frac{6}{x ^{4}} \to - 6$ .
Dividing by $k!$ : $P_{4} = (x - 1) - \frac{( x - 1 ) ^{2}}{2} + \frac{( x - 1 ) ^{3}}{3} - \frac{( x - 1 ) ^{4}}{4}$ .

ln x \approx (x - 1) - \frac{( x - 1 ) ^{2}}{2} + \frac{( x - 1 ) ^{3}}{3} - \frac{( x - 1 ) ^{4}}{4}

Solve it in Simuveliq

In the topic Calculus, operation Taylor series: type ln(x), Center = 1 and Order = 4; press Compute. The Answer is the same polynomial, with the first term written as $x - 1$ at the end.

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Series and Taylor polynomials

Series and partial sums

Convergence tests

Taylor polynomial

Common mistakes

Worked examples

Example 1Geometric series

Solution

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Example 2The harmonic series

Solution

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Example 3A convergent p-series

Solution

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Example 4Conditional convergence

Solution

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Example 5Ratio test

Solution

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Example 6Taylor polynomial of sine

Solution

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Example 7Taylor away from the origin

Solution

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