Why a Power Series Needs a Definition
A polynomial is a finite sum. For example, \(1+2x+3x^2\) has only three terms, so its value at any real \(x\) is determined by ordinary finite addition.
A power series looks similar, but has infinitely many terms. Writing those terms is not by itself enough to say what number the series represents at a given input. We need a precise way to interpret the infinite sum.
The idea is to add the terms one at a time. Each finite sum is called a partial sum. If the sequence of partial sums approaches a finite limit, that limit is the value of the power series at that input. Thus, a power series defines a function only at inputs where this limit exists.
Here \(x\) is the variable: it is the input that may change. The coefficients \(a_n\) are fixed numbers, one for each nonnegative integer \(n\), and \(c\) is a fixed number. The expression \((x-c)^n\) is the \(n\)th power relative to the center.
Partial Sums Give the Infinite Sum Its Meaning
For a fixed input \(x\), the \(N\)th partial sum is obtained by stopping after the term with index \(N\):
This is an ordinary finite sum. The infinite series at \(x\) is defined only if the sequence \(S_0(x),S_1(x),S_2(x),\ldots\) approaches a finite limit as \(N\) increases. When that happens, the notation for the value is
The word “sum” in an infinite series therefore does not mean that infinitely many additions are completed in a final step. It means the limit of a particular sequence of finite sums. If the partial sums do not approach a finite limit, the series does not have a finite value at that input.
Why the Partial-Sum Definition Works
The definition extends finite addition without changing what finite sums mean. For each \(N\), the partial sum contains exactly the first \(N+1\) terms. Increasing \(N\) adds the next term and produces a new finite sum. A limit, if it exists, records the number these finite sums approach.
Choose a particular real value of \(x\). Each expression \(a_n(x-c)^n\) is then a number.
For each \(N\), add the terms from \(n=0\) through \(n=N\) to obtain \(S_N(x)\).
If the partial sums approach a finite number \(L\), define the series value at \(x\) to be \(L\).
The relation between consecutive partial sums makes the construction precise. For \(N\geq 1\), the \(N\)th partial sum consists of the \((N-1)\)st partial sum plus the one new term with index \(N\):
Subtracting \(S_{N-1}(x)\) gives \(S_N(x)-S_{N-1}(x)=a_N(x-c)^N\). This identity follows directly from finite addition: all the earlier terms cancel, leaving precisely the new term. It explains why partial sums, rather than an unstructured “addition of infinitely many terms,” are the objects used to define a series.
Worked Examples
Worked Example: Evaluating a Geometric Power Series at One Input
Consider the power series
Its variable is \(x\), its center is \(c=1\), and its coefficient with index \(n\) is \(a_n=1/2^n\). Evaluate the series at \(x=0\) by using its partial sums.
First substitute \(x=0\) into an individual term:
Thus the \(N\)th partial sum at \(x=0\) is the finite geometric sum
As \(N\) increases, \(\left(-\frac12\right)^{N+1}\) approaches \(0\). Therefore
So the series has value \(2/3\) at \(x=0\). As a finite-sum check, when \(N=2\), direct addition gives \(1-\frac12+\frac14=\frac34\). The formula gives \(\frac23(1-(-\frac12)^3)=\frac23(1+\frac18)=\frac34\), as required.
Worked Example: A Series with Many Zero Terms at a Particular Input
Consider
The variable is \(x\), the center is \(c=-1\), and the coefficient with index \(n\) is \(a_n=n+1\). At the input \(x=-1\), the factor \(x+1\) equals zero. The term with index \(0\) is
because the zeroth power of any base, including zero, is \(1\). For every \(n\geq1\), the term is
Consequently, for every \(N\geq0\), the partial sum is \(S_N(-1)=1+0+\cdots+0=1\). Its limit is \(1\), so the series has value \(1\) at this particular input. This conclusion comes from the partial sums at \(x=-1\); it does not assert anything about other inputs.
Check by substitution in the finite sum: its first term is \(1\), and each remaining term through any chosen index \(N\) is \(0\). Hence every such partial sum is indeed \(1\).
Worked Example: Finding a Value from Telescoping Partial Sums
Consider the power series
Its center is \(c=2\), and its coefficients are \(a_n=1/((n+1)(n+2))\). At \(x=3\), each power \((x-2)^n\) equals \(1^n=1\), so
For each term, partial fractions give
since the right-hand side is \(\frac{(n+2)-(n+1)}{(n+1)(n+2)}=\frac{1}{(n+1)(n+2)}\). Substituting this into the finite sum yields
All intermediate terms cancel in pairs. Since \(1/(N+2)\) approaches \(0\), the limit of the partial sums is \(1\). Thus the series has value \(1\) at \(x=3\).
Check at \(N=2\): direct addition gives \(\frac12+\frac16+\frac1{12}=\frac{6+2+1}{12}=\frac34\). The formula gives \(1-\frac14=\frac34\), confirming the partial-sum expression.
A Common Mistake: Replacing \(x-c\) with \(x\)
The powers in a power series are powers of \(x-c\), not necessarily powers of \(x\). Ignoring the subtraction changes the terms and therefore changes the partial sums.
For example, consider \(\sum_{n=0}^{\infty}2^n(x-3)^n\) at \(x=4\). A mistaken substitution might replace \((x-3)^n\) by \(x^n\), giving the incorrect terms \(2^n4^n=8^n\). The actual substitution is \(x-3=4-3=1\), so the correct \(n\)th term is
For instance, the first four terms, with indices \(0\) through \(3\), are \(1,2,4,8\), and their partial sum is \(1+2+4+8=15\). The incorrect substitution would instead produce \(1+8+64+512=585\). The correction is to substitute the chosen value into the entire factor \(x-c\) before evaluating its powers.
What the Definition Does—and Does Not—Tell You
The definition identifies the structure of a power series and specifies how to interpret its value at an input. It does not, by itself, tell you which inputs make the partial sums converge. That question requires studying the partial sums for the particular coefficients and center. A series may have a finite value at some inputs and fail to have one at others.
It is also useful to keep three objects distinct:
| Object | Meaning |
|---|---|
| Term | The \(n\)th expression \(a_n(x-c)^n\) |
| Partial sum | The finite sum \(S_N(x)=\sum_{n=0}^{N}a_n(x-c)^n\) |
| Series value | The limit of the partial sums, if that finite limit exists |
This distinction is useful whenever an infinite expression appears. First identify the finite approximations. Then determine whether they approach a finite limit before assigning the infinite series a value.
Check Your Understanding
Use the definition and the partial-sum viewpoint to answer these questions.
- In \(\sum_{n=0}^{\infty}a_n(x-c)^n\), which quantities are fixed and which quantity is the variable?
- Write the \(N\)th partial sum of \(\sum_{n=0}^{\infty}(-1)^n(x-2)^n\).
- What condition on the partial sums is required for a power series to have a finite value at a specified input?
- For \(\sum_{n=0}^{\infty}(n+2)(x-4)^n\), what is the term with index \(n=2\)?
- Explain why replacing \(x-c\) by \(x\) can change the value of a partial sum.