Reading the Structure of a Power Series
A power series is an infinite sum whose terms involve powers of a variable. Its notation can look compact, but the details matter: the center determines which expressions are raised to powers, and the coefficients determine what multiplies each power.
The goal here is to read that structure accurately. In particular, we will identify the center and coefficients, expand sigma notation into its first terms, and distinguish the first nonzero terms from terms whose coefficients are zero. These tasks come before deciding where a series converges.
The displayed expression describes the series’ terms. For a particular value of \(x\), the infinite sum represents a number only when it converges at that value. The center and coefficients can still be read from the form of the series without determining its convergence.
Center, Coefficients, and Variable
The center is the number subtracted from the variable inside each power. In \((x-5)^n\), the center is \(5\). In \((x+3)^n\), rewrite the expression as \((x-(-3))^n\): the center is \(-3\), not \(3\).
The coefficient \(c_n\) multiplies the \(n\)th power \((x-a)^n\). The coefficient may depend on the index \(n\), but it does not depend on \(x\). For example, in \(\sum_{n=0}^{\infty}(2n+1)(x-4)^n\), the coefficient at index \(n\) is \(2n+1\), and the variable is \(x\).
A coefficient can be zero. For instance, the expression \(6+0(x-2)+5(x-2)^2\) has coefficient \(c_1=0\). The corresponding power is present in the general pattern, but its term contributes nothing. Therefore, “first terms” and “first nonzero terms” are not always the same list.
The index may start at a value other than zero, and the power may use an index-shifted exponent. To identify the center and initial coefficients reliably, substitute the index’s first values into the actual summand before trying to rewrite it.
Worked Example: Reading a Series Starting at Index Zero
Worked Example: Reading a Series Starting at Index Zero
Identify the center and coefficients of \(\sum_{n=0}^{\infty}(2n+1)(x-4)^n\), and write its first four terms.
The power is \((x-4)^n\), so the center is \(4\). The variable is \(x\). The coefficient at index \(n\) is \(c_n=2n+1\).
Substitute \(n=0\): the coefficient is \(2(0)+1=1\), and the power is \((x-4)^0=1\). This gives \(1\).
Substitute \(n=1\): the coefficient is \(2(1)+1=3\), giving \(3(x-4)\).
Substitute \(n=2\): the coefficient is \(2(2)+1=5\), giving \(5(x-4)^2\).
Substitute \(n=3\): the coefficient is \(2(3)+1=7\), giving \(7(x-4)^3\).
All four displayed coefficients are nonzero, so these are also the first four nonzero terms. As a check on the center, set \(x=4\): every positive power of \(x-4\) becomes zero, and only the constant term \(1\) remains. The center is \(4\), as identified from the powers.
How to Handle a Shifted Index
Suppose a summation begins at \(k=2\), while the exponent is \(k-2\). The first exponent is then \(0\), not \(2\). This distinction is important: the index that labels a term and the exponent on the power need not be the same number.
To express such a series in the standard form with powers numbered \(0,1,2,\ldots\), one can introduce a new index. If \(j=k-2\), then \(k=j+2\). As \(k\) takes the values \(2,3,4,\ldots\), \(j\) takes \(0,1,2,\ldots\). The summand must be rewritten using \(k=j+2\) everywhere, including in its coefficient.
Worked Example: Reindexing a Shifted Series
Find the center and first four terms of \(\sum_{k=2}^{\infty}\frac{(-1)^k}{k}(x+2)^{k-2}\).
First rewrite the power as \((x-(-2))^{k-2}\). The center is \(-2\), and the variable is \(x\). To list the terms, substitute the first values of \(k\) into both the coefficient and exponent.
For \(k=2\), the coefficient is \(\frac{(-1)^2}{2}=\frac12\), and the exponent is \(2-2=0\). The term is \(\frac12\).
For \(k=3\), the coefficient is \(\frac{(-1)^3}{3}=-\frac13\), and the exponent is \(3-2=1\). The term is \(-\frac13(x+2)\).
For \(k=4\), the coefficient is \(\frac{(-1)^4}{4}=\frac14\), and the exponent is \(4-2=2\). The term is \(\frac14(x+2)^2\).
For \(k=5\), the coefficient is \(\frac{(-1)^5}{5}=-\frac15\), and the exponent is \(5-2=3\). The term is \(-\frac15(x+2)^3\).
Equivalently, putting \(j=k-2\) gives \(k=j+2\), so the coefficient becomes \(\frac{(-1)^{j+2}}{j+2}=\frac{(-1)^j}{j+2}\), and the exponent becomes \(j\). This agrees with the four terms above. At \(x=-2\), all positive powers vanish and the value is the constant term \(\frac12\), which checks the center and the first coefficient.
First Nonzero Terms in an Expanded Series
An expanded expression may show zero terms explicitly, or it may omit them. Read each coefficient alongside its matching power. Do not assume that the first term shown with a positive exponent is the first nonzero term, and do not assume that the coefficient of every successive power is nonzero.
For example, a series written as \(4+0(x-1)-2(x-1)^2+\cdots\) has coefficient zero for the first power. Its first three nonzero terms begin with the constant \(4\), followed by the quadratic term \(-2(x-1)^2\); a linear term does not appear among them.
Worked Example: Finding Nonzero Terms in an Expanded Series
Consider the expanded series \[ 2+0(x-1)-3(x-1)^2+\frac{(x-1)^3}{3!}+\frac{(x-1)^4}{4!}+\cdots, \] where the terms from degree \(3\) onward follow the pattern \(\frac{(x-1)^n}{n!}\) for \(n\geq 3\). Identify its center and first four nonzero terms.
The powers are powers of \(x-1=x-1\), so the center is \(1\). The variable is \(x\). The constant term is the degree-zero term, so \(c_0=2\). The degree-one coefficient is \(c_1=0\).
The degree-two term is \(-3(x-1)^2\), so \(c_2=-3\). The degree-three coefficient is \(\frac{1}{3!}=\frac16\), and the degree-four coefficient is \(\frac{1}{4!}=\frac{1}{24}\).
The degree-one term is zero and must be skipped when listing nonzero terms. Thus the first four nonzero terms are
For a check, at \(x=1\) each term of positive degree is zero, including the degree-one term whose coefficient is already zero. The series at the center therefore reduces to its constant term, \(2\), consistent with the identified center and \(c_0\).
Why the Center Is the Natural Reference Point
The center is more than a label. It is the input value that makes the powers easy to distinguish: at \(x=a\), the base \(x-a\) equals zero. The constant term has exponent zero, while each term of positive degree contains a positive power of zero.
Write its partial sum through degree \(N\) as \(S_N(x)=\sum_{n=0}^{N}c_n(x-a)^n\).
At \(x=a\), the factor \(x-a\) is zero, so \(S_N(a)=c_0+\sum_{n=1}^{N}c_n0^n=c_0\).
Every partial sum at \(a\) equals the same number \(c_0\). Therefore, the series at its center converges to \(c_0\).
This is a complete reason that the constant coefficient gives the value at the center: no matter how many terms are included, every positive-degree term vanishes there. The calculation does not say what happens at other values of \(x\); that is a separate question about convergence.
A Common Indexing Mistake
A frequent error is to treat the lower limit of the summation as though it were automatically the exponent. In the shifted-index example, the series begins at \(k=2\), but the first exponent is \(k-2=0\). Writing the first term as \(\frac12(x+2)^2\) confuses the index with the exponent.
The correction is to substitute \(k=2\) into the exponent itself: \(2-2=0\). The first term is \(\frac{(-1)^2}{2}(x+2)^0=\frac12\). A second common error is to call the center \(2\) because the series contains \(x+2\). Since \(x+2=x-(-2)\), the center is \(-2\).
A Reliable Reading Procedure
When you meet a power series in either sigma notation or expanded form, use a consistent sequence of checks:
- Locate the variable and identify the repeated power expression.
- Rewrite that expression as a power of \(x-a\) to read the center \(a\).
- Find the coefficient attached to each power; keep index substitutions consistent.
- For sigma notation, substitute the first index values to generate terms in order.
- Skip terms with zero coefficients when asked for nonzero terms.
- Check the center by setting \(x=a\): only the constant term remains.
These steps describe the anatomy of a power series without making a claim about its behavior away from the center. In particular, identifying its structure is not the same as finding where the series converges.
Check Your Understanding
For each question, identify the requested feature from the series’ structure. You do not need to determine any convergence interval.
- For \(\sum_{n=0}^{\infty}(-2)^n(x-3)^n\), what are the center and the coefficient \(c_2\)?
- Write the first three terms of \(\sum_{k=1}^{\infty}\frac{1}{k+1}(x+4)^{k-1}\), and identify its center.
- In \(5+0(x-2)+7(x-2)^3+\cdots\), what is the coefficient of \((x-2)^2\), and what are the first two nonzero terms shown?
- At what value of \(x\) does \(\sum_{n=0}^{\infty}c_n(x-6)^n\) reduce to its constant term, and what is that term?
- For \(\sum_{k=3}^{\infty}a_k(x-1)^{k-3}\), what exponent does the first term have? What is the center?
