Why the First Omitted Term Can Control an Error
A partial sum of an infinite series is an approximation to its full sum. When the series alternates, successive terms tend to offset one another. If the term magnitudes decrease to zero, this cancellation gives a useful and simple error estimate: the error after truncation cannot exceed the magnitude of the first term left out.
This result is especially useful with Taylor series. It can turn a truncated Taylor polynomial into a certified approximation, provided the series actually represents the function at the point in question and its terms meet the required conditions. The bound is not automatic for every alternating-looking expression; both the behavior of the terms and the meaning of the series must be checked.
Definition and Remainder Bound
The alternating-series remainder bound applies when these magnitudes are nonincreasing and tend to zero. “Nonincreasing” allows equal successive magnitudes; strict decrease is not required.
The same conclusion holds if the first term has a minus sign, since changing the sign of the whole series does not change the magnitude of its remainder. In practice, it is often clearer to count included terms directly: identify the next term in the original series, then bound the error by that term’s magnitude.
Why the Bound Holds
First consider the series with a positive first term. Its partial sums with even indices decrease, because \(S_{2k+2}-S_{2k}=-b_{2k+1}+b_{2k+2}\le 0\). The partial sums with odd indices increase, because \(S_{2k+1}-S_{2k-1}=b_{2k}-b_{2k+1}\ge 0\). Also \(S_{2k}-S_{2k-1}=b_{2k}\ge 0\), and this gap tends to zero. The even and odd subsequences therefore approach the same limit, so the full sequence of partial sums converges.
Now look at a finite piece of the tail, beginning with the first omitted term, and factor out its sign. Its magnitude has the form \(T_q=b_{m+1}-b_{m+2}+b_{m+3}-\cdots+(-1)^{q-1}b_{m+q}\). Pairing neighboring terms shows \(T_q\ge 0\): each pair \(b_{m+2j+1}-b_{m+2j+2}\) is nonnegative, with a final nonnegative term if \(q\) is odd. Also \(T_q\le b_{m+1}\). To see this, subtract \(T_q\) from \(b_{m+1}\); the result is another sum of nonnegative differences, with a possible final nonnegative term. Hence every such finite tail lies between zero and \(b_{m+1}\). Taking the limit as the tail length grows gives \(|R_m|\le b_{m+1}\), as claimed.
This argument also establishes convergence under the stated conditions. The error estimate is not based on guessing how much cancellation occurs: it follows from the ordering and limiting behavior of the term magnitudes.
Applying the Bound to Taylor Series
Suppose a Taylor series for \(f(x)\), centered at \(a\), converges to \(f(x)\) at the chosen value of \(x\). If its terms at that value alternate in sign, their magnitudes decrease, and those magnitudes tend to zero, then truncating the series gives an approximation whose error is no greater than the first omitted term’s magnitude.
There are two distinct checks here. First, the sequence of Taylor-term magnitudes must satisfy the alternating-series conditions. Second, the Taylor series must equal the function at the point being approximated. A Taylor series is not guaranteed to represent its function everywhere merely because its coefficients come from derivatives; the earlier material on Taylor series addresses this distinction.
Worked Example: Approximating \(\ln(3/2)\)
For \(|x|<1\), the logarithm series is \(\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\). At \(x=\frac12\), it becomes \(\ln(3/2)=\frac12-\frac{(1/2)^2}{2}+\frac{(1/2)^3}{3}-\frac{(1/2)^4}{4}+\cdots\).
Include the first four terms. Their magnitudes are \(b_n=(1/2)^n/n\), for \(n\ge1\). For successive terms, \(b_{n+1}/b_n=\frac12\frac{n}{n+1}<1\), so the magnitudes decrease; they also tend to zero. The series represents the logarithm because \(|1/2|<1\).
The four-term approximation is \[ \frac12-\frac18+\frac1{24}-\frac1{64} =\frac{96-24+8-3}{192} =\frac{77}{192}. \] The first omitted term is positive and has magnitude \[ \frac{(1/2)^5}{5}=\frac{1}{32\cdot5}=\frac1{160}. \] Therefore, \[ \left|\ln(3/2)-\frac{77}{192}\right|\le\frac1{160}. \]
Check the indexing: the included terms have powers \(1,2,3,4\), so the first omitted term has power \(5\), not power \(4\). The bound is therefore \(1/160\). The fraction for the approximation also checks directly: \(96-24+8-3=77\), all over \(192\).
Worked Example: Approximating \(\arctan(1/2)\)
For \(|x|<1\), the Taylor series is \(\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots\). At \(x=\frac12\), use three terms: \[ \arctan(1/2)\approx\frac12-\frac{(1/2)^3}{3}+\frac{(1/2)^5}{5} =\frac12-\frac1{24}+\frac1{160}. \]
With denominator \(480\), the sum is \[ \frac{240-20+3}{480}=\frac{223}{480}. \] For the magnitude of the term indexed by \(n\ge0\), \(b_n=x^{2n+1}/(2n+1)\). At \(x=1/2\), \[ \frac{b_{n+1}}{b_n} =\frac14\frac{2n+1}{2n+3}<1, \] and \(b_n\) tends to zero. Since \(|1/2|<1\), the series represents \(\arctan(1/2)\). The first omitted term is \(-\frac{(1/2)^7}{7}\), whose magnitude is \(1/(128\cdot7)=1/896\). Thus \[ \left|\arctan(1/2)-\frac{223}{480}\right|\le\frac1{896}. \]
The arithmetic and the error index agree: three terms use powers \(1,3,5\); the next power is \(7\), giving the stated bound.
Worked Example: Approximating \(\cos(1)\)
The Taylor series for cosine gives \(\cos(1)=1-\frac1{2!}+\frac1{4!}-\frac1{6!}+\cdots\). Using the first three terms, \[ 1-\frac1{2!}+\frac1{4!} =1-\frac12+\frac1{24} =\frac{24-12+1}{24} =\frac{13}{24}. \]
The term magnitudes are \(b_n=1/(2n)!\). Their successive ratio is \[ \frac{b_{n+1}}{b_n} =\frac{1}{(2n+2)(2n+1)}<1, \] and \(b_n\) tends to zero. The cosine Taylor series represents \(\cos(1)\). After three terms, the next term is \(-1/6!=-1/720\), so \[ \left|\cos(1)-\frac{13}{24}\right|\le\frac1{720}. \]
As a check on the truncation, the included factorial denominators are \(0!,2!,4!\); the first omitted denominator is \(6!\), with \(6!=720\). Thus the bound uses the first term not included in the approximation.
Conditions and Common Pitfalls
The most frequent indexing error is to use the last included term rather than the first omitted one. If an approximation contains four terms, the relevant bound comes from term five. Count the actual terms in the series, especially when it begins at \(n=0\) or when only even powers appear.
Another mistake is to assume that alternating signs alone are enough. For example, the series with terms \(1-2+1-2+\cdots\) alternates, but its term magnitudes do not decrease to zero. The theorem does not apply. A proposed bound based on its next term would have no justification from this result.
Finally, do not confuse a Taylor polynomial with a guaranteed approximation to the function. The remainder bound estimates the difference between a partial sum and the sum of the infinite series. It estimates the difference between the partial sum and \(f(x)\) only when that series is known to converge to \(f(x)\) at the point in question.
Check Your Understanding
For each question, identify which condition or term of the remainder bound is relevant.
- A partial sum includes terms through index \(m=5\). Which term magnitude supplies the alternating-series error bound?
- Why does an alternating series whose term magnitudes fail to tend to zero not meet the theorem’s hypotheses?
- For \(\ln(1+x)\) at \(x=1/3\), what is the magnitude of the first omitted term after using terms through \(x^4\)?
- What additional fact, beyond the alternating-series conditions on its terms, is needed to use the bound for a Taylor approximation to \(f(x)\)?
- In the cosine example, why is \(1/720\), rather than \(1/24\), the error bound after three terms?
