Why checking the conditions matters
A first-omitted-term error bound can make a Taylor approximation useful: it controls the difference between a function value and a partial sum using just one term. But the bound is not automatic whenever Taylor terms have alternating signs. Their magnitudes must also decrease along the tail, and must tend to zero. Checking these conditions at the particular input value is essential.
Sometimes the magnitudes do not decrease from the first term, but do decrease after a finite number of terms. In that case, the alternating-series bound can still apply to a tail that starts late enough. A Taylor polynomial must include all earlier terms before that tail begins.
The conditions for a Taylor-series tail
For a function \(f\) with a Taylor series centered at \(a\), write the term of degree \(k\), evaluated at \(x\), as
Here \(x\) is fixed: the checks concern the numerical sequence of terms at that input, not the functions \(t_k\) over an interval. The Taylor polynomial through degree \(n\) is \(T_n(x)=\sum_{k=0}^{n}t_k(x)\). If the Taylor series converges to \(f(x)\), its remainder is the tail \(f(x)-T_n(x)=\sum_{k=n+1}^{\infty}t_k(x)\).
“Nonincreasing” allows equal consecutive magnitudes. In practice, sign alternation can often be established by writing \(t_k(x)=(-1)^k b_k\), or \(t_k(x)=(-1)^{k+1}b_k\), with \(b_k\geq 0\). A fixed overall negative sign changes which sign comes first, but not whether the signs alternate.
Why the first-omitted-term bound holds
It is useful to see why the conditions lead to a bound. Consider a tail whose first term is positive, and denote its magnitudes by \(c_1,c_2,\ldots\), where \(c_j\geq c_{j+1}\) and \(c_j\to 0\). Its partial sums have the form \(s_r=c_1-c_2+c_3-\cdots+(-1)^{r+1}c_r\).
For each \(m\), \(s_{2m}=(c_1-c_2)+\cdots+(c_{2m-1}-c_{2m})\geq 0\). Also, \(s_{2m+2}-s_{2m}=c_{2m+1}-c_{2m+2}\geq 0\), so the even partial sums are nondecreasing.
Since \(s_{2m}=c_1-(c_2-c_3)-\cdots-(c_{2m-2}-c_{2m-1})-c_{2m}\leq c_1\), they are bounded above by \(c_1\).
For every \(m\), \(s_{2m+1}=s_{2m}+c_{2m+1}\). Since \(c_{2m+1}\to 0\), the odd and even partial sums have the same limit. Thus the tail converges to some value \(S\) with \(0\leq S\leq c_1\).
If the first term is negative, multiplying the tail by \(-1\) makes it positive without changing its absolute value. Therefore, in either case, the magnitude of the tail is at most \(c_1\), the magnitude of its first term. The argument applies to any tail that satisfies the conditions, even if earlier Taylor terms did not.
Worked Example: The logarithm series at an endpoint
Worked Example: The logarithm series at an endpoint
Consider the Taylor series for \(\ln(1+x)\) at \(x=0\), evaluated at \(x=1\), and approximate \(\ln 2\) by the polynomial through degree \(3\).
The terms at \(x=1\), starting with \(k=1\), are \(t_k(1)=(-1)^{k+1}/k\). Their signs alternate, and their magnitudes are \(b_k=1/k\). For every \(k\geq 1\), \(b_{k+1}=1/(k+1)<1/k=b_k\), and \(\lim_{k\to\infty}b_k=0\). The Taylor series at this endpoint sums to \(\ln 2\), so the alternating-series bound applies.
The degree-three polynomial is
The first omitted term is \(t_4(1)=-1/4\), with magnitude \(1/4\). Therefore,
As a check, the tail starts with \(-1/4\), so the error has that sign or is zero; its magnitude cannot exceed \(1/4\). The first omitted term and the bound agree, and all three conditions hold for the tail.
When the conditions hold only eventually
To say that magnitudes decrease eventually means there is an index \(K\) such that \(b_{k+1}\leq b_k\) for every \(k\geq K\). It does not mean the sequence is decreasing from its first term. If signs alternate and the magnitudes tend to zero, the bound is available for a remainder whose first omitted index is at least \(K\). It is not justified for an earlier truncation merely because the sequence decreases later.
Worked Example: An exponential tail that decreases only after some terms
Consider the Taylor series for \(e^{-3}\), and use its degree-three polynomial. Its terms are \(t_k=(-3)^k/k!\), so their signs alternate and their magnitudes are \(b_k=3^k/k!\). To test monotonicity, compute the ratio of consecutive magnitudes:
This ratio is greater than \(1\) at \(k=0,1\), equal to \(1\) at \(k=2\), and less than \(1\) for \(k\geq 3\). Thus the magnitudes are not decreasing from the start, but are nonincreasing from \(k=2\) onward. Also \(b_k\to 0\): for \(k\geq 5\), the ratio \(3/(k+1)\leq 1/2\), so each later magnitude is at most half the preceding one.
The polynomial through degree three is
The first omitted term has index \(4\), which is in the decreasing part of the sequence:
The Taylor series for the exponential converges to \(e^{-3}\). The alternating-series bound therefore gives \(|e^{-3}-(-2)|\leq 27/8\). The tail starts with a positive term, and its magnitude is at most \(27/8\), as required. Although the early magnitudes increase, they are already included in \(T_3\); the omitted tail starts only after the conditions hold.
Checking a failed case instead of forcing the bound
A calculation that produces alternating signs is not enough. If the terms do not tend to zero, the alternating-series test fails, so there is no first-omitted-term error bound from that test. A Taylor series can also fail to represent the function at the chosen input, even if a sequence of terms appears alternating. Always check the series’ convergence to the function as well as the tail conditions.
Worked Example: A geometric Taylor series outside its convergence interval
For \(f(x)=1/(1+x)\), the Taylor terms at \(x=0\) are \(t_k(x)=(-1)^k x^k\). At \(x=2\), the terms are \(t_k(2)=(-2)^k\), with magnitudes \(b_k=2^k\). The signs alternate, but the magnitudes increase: \(b_{k+1}/b_k=2\). They also fail to tend to zero; in fact, \(2^k\) grows without bound. Therefore, the alternating-series error bound cannot be used.
For example, the degree-two Taylor polynomial at \(x=2\) is \(T_2(2)=1-2+4=3\), while \(f(2)=1/3\). The actual difference is
The first omitted term has magnitude \(|t_3(2)|=8\), but that number is not a justified error bound: the conditions fail, and the Taylor series does not converge to \(f(2)\). The fact that \(8\) happens to exceed this particular error does not validate the method. The checks reject the bound, which is the correct conclusion.
A practical verification routine
For a specific Taylor approximation, carry out the checks in an order that matches the remainder you need to control:
- Fix the input \(x\) and write the Taylor term \(t_k(x)\) explicitly.
- Check the signs of the terms in the omitted tail. If there are zero terms, inspect the nonzero terms and confirm that their signs alternate in order.
- Set \(b_k=|t_k(x)|\). Establish \(b_{k+1}\leq b_k\) for every index in the tail, not just for a few initial comparisons.
- Show that \(b_k\to 0\). A ratio estimate can help, but its conclusion must hold for all sufficiently large indices.
- Confirm that the Taylor series converges to the function value at the chosen input.
- Only then compare the error with the magnitude of the first omitted term.
For factorial terms, the ratio \(b_{k+1}/b_k\) often reduces to a simple expression in \(k\). If that ratio is at most \(1\) from some index onward, the magnitudes are nonincreasing from that point. The first omitted index must be at or beyond that point. When the ratio is strictly less than \(1\) eventually, it can also help establish that the magnitudes tend to zero.
Common pitfall: checking only the first pair
It is a mistake to compare the first omitted term with the next one, see that the first is larger, and conclude that the entire tail decreases. The required condition applies to every successive pair from the first omitted term onward. A ratio formula, or a general inequality valid for all later indices, supplies the needed verification. Likewise, checking that magnitudes eventually decrease does not justify a bound for a truncation whose tail starts before that decrease begins.
Check Your Understanding
For each question, identify which conditions must be checked before a first-omitted-term bound can be claimed.
- A Taylor tail has alternating signs and decreasing magnitudes, but the magnitudes do not tend to zero. Is the alternating-series error bound justified? Explain.
- The term magnitudes increase for the first three indices and decrease thereafter. What must be true of the truncation index for an eventual alternating-series bound to apply?
- For \(b_k=5^k/k!\), compute \(b_{k+1}/b_k\) and state the condition on \(k\) that makes the magnitudes decrease.
- Why must one check that the Taylor series sums to \(f(x)\), rather than only proving that the series of terms converges?
- If the alternating-series conditions hold for the omitted tail beginning with \(t_{n+1}(x)\), what quantity bounds the absolute approximation error?