Write the requested terms or general term
From the terms to one formula
The table lists the series \(x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots\). Each column is one term. The index \(n\) counts the terms, and the first term has \(n=0\).
Press the buttons to light a column in orange. The power of \(x\) is 1, 3, 5, 7, which is \(2n+1\). The denominator equals the power, so it is \(2n+1\) too.
The sign alternates between a green + and a red −. The factor \((-1)^n\) gives this, because it is 1 for even \(n\) and −1 for odd \(n\).
Press Show the formulas. The three formulas combine into one general term, \(\frac{(-1)^n x^{2n+1}}{2n+1}\).
Three formulas and a start
The table has three rows, the sign, the power of x and the denominator, and each row holds a formula in n.
Press Start at n = 1. Because the first term now has \(n=1\), every \(n\) in the formulas is replaced by \(n-1\), and the terms of the series stay the same.
A formula is correct when it gives the listed terms for the first values of n.
Notation
Write the first few terms. Number them with an index \(n\), and choose where \(n\) starts. Here \(n\) starts at 1.
Find a formula in n for each part of the term: the sign, the power of x and the denominator. Here the power and the denominator are both n.
If the signs alternate and the first term is positive, the sign factor depends on where \(n\) starts.
\[\begin{aligned} (-1)^n&\ \text{for } n=0,1,2,\dots \\ (-1)^{n-1}&\ \text{for } n=1,2,3,\dots \end{aligned}\]The first term is positive, so the exponent of \(-1\) must be even at the first value of \(n\).
If the denominator has a factorial, take its pattern from the standard series. The number in the factorial is not always the power of \(x\).
\[\sin x:\ \dfrac{(-1)^n\,x^{2n+1}}{(2n+1)!}\]Check the formula. Substitute \(n=1,2,3\) and compare the terms with the list.
If the question asks for the first nonzero terms, write the series in order of power and skip every term that is 0.
When it fails
In the series \(x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\), the denominator is \(n\). At \(n=0\) the denominator is 0, so that term does not exist.
The sum must start at \(n=1\). The first term \(x\) is positive, so the sign is \((-1)^{n-1}\), and the general term is \(\frac{(-1)^{n-1}x^n}{n}\).
Example 1: a pattern in the denominators
Write the general term of \(\dfrac12+\dfrac{x}{4}+\dfrac{x^2}{8}+\dfrac{x^3}{16}+\cdots\), with \(n\) starting at 0.
The powers of \(x\) are 0, 1, 2, 3. The power in term \(n\) is \(n\).
\[x^{n}\]The denominators are 2, 4, 8, 16. Each one is a power of 2, and the exponent is one more than \(n\), so the denominator in term \(n\) is \(2^{n+1}\).
\[2^{n+1}\]All the signs are positive, so no sign factor is needed. The general term is the power over the denominator.
\[\dfrac{x^{n}}{2^{n+1}}\]
Example 2: a different center
Write the general term of the Taylor series of \(e^{x}\) about \(x=3\).
A Taylor series about 3 uses powers of \(x-3\). The coefficient of the \(n\)-th power is \(f^{(n)}(3)/n!\), and every derivative of \(e^x\) is \(e^x\), so every \(f^{(n)}(3)\) is \(e^3\). The orange curve shows the first three terms.
\[f^{(n)}(3)=e^{3}\]A tempting wrong move: The coefficients are \(e^3/n!\), so the series is the sum of \(e^3x^n/n!\).
\[\sum_{n=0}^{\infty}\dfrac{e^{3}\,x^{n}}{n!}\]At \(x=0\), \(e^x\) equals 1, but this series, the red curve, equals \(e^3\approx 20.1\) there. The powers must be powers of \(x-3\), because the center is 3.
Use powers of \(x-3\). The first three terms are \(e^3\), \(e^3(x-3)\) and \(\frac{e^3(x-3)^2}{2}\), and the orange curve touches the chalk curve at the center.
\[e^{3}\sum_{n=0}^{\infty}\dfrac{(x-3)^{n}}{n!}\]
Practice
Write the general term of \(x^3-\dfrac{x^5}{3!}+\dfrac{x^7}{5!}-\cdots\), with \(n\) starting at 0.
Compare the power of x with the number in the factorial. The difference is the same in every term.
Solution
The powers are 3, 5, 7. The power in term \(n\) is \(2n+3\).
\[x^{2n+3}\]The factorials are \(1!\), \(3!\), \(5!\). The factorial in term \(n\) is \((2n+1)!\). It is 2 less than the power, because this series is \(x^2\) times the sine series.
\[(2n+1)!\]The signs are +, −, +, so the sign is \((-1)^n\).
\[\dfrac{(-1)^n\,x^{2n+3}}{(2n+1)!}\]
Questions
The table shows the terms of the series \(1-\frac{3x^2}{2!}+\frac{9x^4}{4!}-\frac{27x^6}{6!}+\cdots\). Each column is one term, and the first column has \(n=0\).
Which formula is the general term, with n starting at 0?
- \(\dfrac{(-3)^n x^{2n}}{(2n)!}\)
- \(\dfrac{(-3)^n x^{2n}}{n!}\)
- \(\dfrac{(-3)^n x^{n}}{(2n)!}\)
- \(\dfrac{3^n x^{2n}}{(2n)!}\)
Answer
\(\dfrac{(-3)^n x^{2n}}{(2n)!}\)
The same series is written with n starting at 1. Which formula is the general term?
- \(\dfrac{(-3)^{n-1} x^{2n-2}}{(2n-2)!}\)
- \(\dfrac{(-3)^{n} x^{2n-2}}{(2n-2)!}\)
- \(\dfrac{(-3)^{n-1} x^{2n}}{(2n)!}\)
- \(\dfrac{(-3)^{n-1} x^{2n-2}}{(2n)!}\)
Answer
\(\dfrac{(-3)^{n-1} x^{2n-2}}{(2n-2)!}\)
Which formula is the general term of the series for \(x^2e^{-x}\), with \(n\) starting at 0?
- \(\dfrac{(-1)^n x^{n+2}}{n!}\)
- \(\dfrac{(-1)^n x^{n+2}}{(n+2)!}\)
- \(\dfrac{(-1)^n x^{2n}}{n!}\)
- \(\dfrac{x^{n+2}}{n!}\)
Answer
\(\dfrac{(-1)^n x^{n+2}}{n!}\)
Summary
You write the sign, the power of x and the denominator of the first terms as formulas in n, and you choose where n starts.
The formulas are correct because they give the listed terms when you substitute the first values of n.
You use this to write a series in sigma notation or to list its first nonzero terms.
This is one of four skills in topic 10.14 of Which & Why: AP Calculus.