Error Bounds for Logarithmic Series

Why logarithmic error bounds need attention

A logarithmic power series can give a useful numerical approximation, but its terms do not all behave the same way across the interval of convergence. For positive $x$, the series for $\ln(1+x)$ alternates, so the first omitted term gives a simple error bound. For negative $x$, its terms all have the same sign; the alternating-series estimate no longer applies. In either case, the location of $x$ relative to $-1$, where the logarithm is undefined, matters.

The goal is not just to produce a partial sum. A useful estimate states how far that sum can be from the exact value and explains why the estimate is valid.

Definition: If $P_N(x)$ is an approximation to $f(x)$, its truncation error is $R_N(x)=f(x)-P_N(x)$. An error bound is a proven inequality of the form $|R_N(x)|\leq B$, where $B$ is a known nonnegative quantity.

The logarithmic series and its interval

For $-1<x\leq 1$, the power series for the natural logarithm is

\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n} =x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots.

For $|x|<1$, this follows by integrating the geometric series for $1/(1+x)$. At $x=1$, the series converges to $\ln 2$, and the alternating-series estimate still applies. At $x=-1$, $\ln(1+x)$ is undefined and the series does not converge. These endpoint conditions are important: a bound that uses this expansion cannot simply be carried across $x=-1$.

Let $P_N(x)$ denote the sum through the $x^N$ term:

P_N(x)=\sum_{n=1}^{N}(-1)^{n+1}\frac{x^n}{n}.

When $0<x\leq 1$, the magnitudes $x^n/n$ decrease to zero. The alternating-series remainder estimate therefore gives an error no larger than the first omitted term. In fact, for this logarithmic series the remainder also has a known sign.

Theorem: If $0<x\leq 1$, then $$ (-1)^N\bigl(\ln(1+x)-P_N(x)\bigr)\geq 0 \qquad\text{and}\qquad \left|\ln(1+x)-P_N(x)\right|\leq\frac{x^{N+1}}{N+1}. $$

Why the logarithmic remainder has this bound

A finite geometric identity gives a direct justification, including the sign of the error. For $t\neq -1$,

\frac{1}{1+t} =\sum_{k=0}^{N-1}(-t)^k+\frac{(-t)^N}{1+t}.

Integrate both sides from $0$ to $x$, where $0<x\leq1$. The left side integrates to $\ln(1+x)$. The finite sum integrates term by term to $P_N(x)$. Thus the remainder has the exact integral form

\ln(1+x)-P_N(x) =(-1)^N\int_0^x\frac{t^N}{1+t}\,dt.

On this interval, $t^N/(1+t)\geq0$. The integral therefore has sign $(-1)^N$, which establishes the sign statement. Also $1+t\geq1$, so

\left|\ln(1+x)-P_N(x)\right| =\int_0^x\frac{t^N}{1+t}\,dt \leq\int_0^x t^N\,dt =\frac{x^{N+1}}{N+1}.

This proves the error bound, rather than merely suggesting it from the size of the next term. The same estimate is valid at $x=1$: the integral is finite, and the upper bound $1/(N+1)$ tends to zero as $N$ increases.

Worked Example: Approximating $\ln(1.2)$

Use the first three terms of the series and bound the truncation error. Here $x=0.2=1/5$ and $N=3$, so the partial sum is

P_3(0.2)=0.2-\frac{(0.2)^2}{2}+\frac{(0.2)^3}{3}.

Compute each term:

0.2=\frac15,\qquad \frac{(0.2)^2}{2}=\frac{0.04}{2}=0.02,\qquad \frac{(0.2)^3}{3}=\frac{0.008}{3}=0.002666\ldots.

Therefore $P_3(0.2)=0.182666\ldots$. The first omitted term has magnitude

\frac{(0.2)^4}{4}=\frac{0.0016}{4}=0.0004.

Since $N=3$, the remainder has sign $(-1)^3$, so it is nonpositive. The estimate is

0.182266\ldots =P_3(0.2)-0.0004 \leq\ln(1.2)\leq P_3(0.2) =0.182666\ldots.

The claimed error is at most $0.0004$. As a check, the next partial sum is $P_4(0.2)=0.182666\ldots-0.0004=0.182266\ldots$; the alternating-series estimate places the exact value between $P_4$ and $P_3$, as shown.

Negative inputs: use the interval, not the alternating rule

If $-1<x<0$, every term of the expansion for $\ln(1+x)$ is negative. It is incorrect to apply the alternating-series estimate in this case. One option is Taylor’s theorem: if $f$ has $N+1$ continuous derivatives between $0$ and $x$, then

f(x)=P_N(x)+R_N(x), \qquad |R_N(x)|\leq\frac{M|x|^{N+1}}{(N+1)!},

provided $|f^{(N+1)}(u)|\leq M$ throughout that interval. For $f(x)=\ln(1+x)$,

f^{(N+1)}(u)=\frac{(-1)^N N!}{(1+u)^{N+1}}.

On an interval $[-a,0]$ with $0<a<1$, the denominator satisfies $1+u\geq1-a>0$. Thus one can take $M=N!/(1-a)^{N+1}$. The bound grows as the interval approaches $-1$, reflecting the logarithm’s singularity there.

Worked Example: Bounding the error for $\ln(0.8)$

Since $0.8=1-0.2$, use $x=-0.2$ and the degree-four Taylor polynomial:

P_4(-0.2) =-0.2-\frac{(0.2)^2}{2} -\frac{(0.2)^3}{3} -\frac{(0.2)^4}{4}.

The terms are $-0.2$, $-0.02$, $-0.002666\ldots$, and $-0.0004$. Hence $P_4(-0.2)=-0.223066\ldots$.

For $u\in[-0.2,0]$, $1+u\geq0.8$. The fifth derivative is $f^{(5)}(u)=24/(1+u)^5$, so $|f^{(5)}(u)|\leq24/(0.8)^5$. Taylor’s theorem gives

|R_4(-0.2)| \leq \frac{24}{(0.8)^5}\frac{(0.2)^5}{5!} =\frac{(0.2)^5}{5(0.8)^5} =\frac{1}{5120} =0.0001953125.

Therefore $\ln(0.8)$ lies within $0.0001953125$ of $-0.223066\ldots$. The derivative bound is valid because the entire interval from $-0.2$ to $0$ stays to the right of $-1$. As a check on the scale and direction, the next series term is $-(0.2)^5/5=-0.000064$, which has magnitude less than the Taylor bound and is consistent with the negative remainder.

A transformed series for logarithms

When a positive number $y$ is not close to $1$, it can be helpful to set $z=(y-1)/(y+1)$. For $y>0$, this gives $|z|<1$ and $y=(1+z)/(1-z)$. The geometric-series expansion of $1/(1-z^2)$, integrated from $0$ to $z$, yields

\ln y =\ln\left(\frac{1+z}{1-z}\right) =2\left(z+\frac{z^3}{3}+\frac{z^5}{5}+\cdots\right).

For $0<z<1$, these terms are positive rather than alternating. If the sum is retained through $z^{2m+1}/(2m+1)$, the remaining terms satisfy

0<R_m =2\sum_{k=m+1}^{\infty}\frac{z^{2k+1}}{2k+1} \leq \frac{2z^{2m+3}}{(2m+3)(1-z^2)}.

To justify the inequality, every denominator $2k+1$ in the tail is at least $2m+3$. Replacing each by this smaller denominator can only increase the sum. The remaining powers form a geometric series with sum $z^{2m+3}/(1-z^2)$.

Worked Example: Bounding an approximation to $\ln 3$

Here $z=(3-1)/(3+1)=1/2$. Keep four terms, through $z^7/7$:

A =2\left(\frac12+\frac{(1/2)^3}{3} +\frac{(1/2)^5}{5}+\frac{(1/2)^7}{7}\right) =2\left(\frac12+\frac1{24}+\frac1{160}+\frac1{896}\right).

Using common denominator $13440$ inside the parentheses gives

\frac12+\frac1{24}+\frac1{160}+\frac1{896} =\frac{6720+560+84+15}{13440} =\frac{7379}{13440}.

Thus $A=7379/6720=1.098065\ldots$. The tail bound, with $m=3$ and $z=1/2$, is

0<\ln 3-A \leq \frac{2(1/2)^9}{9(1-(1/2)^2)} =\frac{1}{1728} =0.000578\ldots.

So $1.098065\ldots<\ln 3\leq1.098065\ldots+0.000578\ldots$. The direction is correct because every omitted term is positive. The geometric tail estimate applies because $z=1/2$ is strictly between $0$ and $1$.

Common mistake: treating every next term as an error bound

The “first omitted term” rule is not universal. It applies to a convergent alternating series whose term magnitudes decrease to zero, not automatically to every power series. For example, at $x=-0.2$, the series for $\ln(1+x)$ has consecutive negative terms. Saying “the error is at most the next term by the alternating-series estimate” gives a conclusion without satisfying the estimate’s alternating-sign hypothesis.

The correction is to check the method’s conditions. For a negative input, use a Taylor remainder bound on an interval where the derivative is bounded, or derive a suitable bound for the nonalternating tail. For a positive input $0<x\leq1$, the logarithmic terms do alternate and decrease, so the first omitted term does provide a valid bound. For a transformed series with positive terms, bound the tail using a geometric comparison, as in the $\ln 3$ example.

Takeaway: Before using an error estimate, check the input interval, the signs and sizes of the omitted terms, and any derivative bound the method requires. A number that resembles the next term is not an error bound until those conditions are justified.

Check Your Understanding

For each question, identify the approximation and the conditions that make the proposed error estimate valid.

For $P_4(0.4)$, what is the first omitted term’s magnitude, and what does the alternating-series estimate say about the remainder?
Why can the alternating-series estimate not be applied directly to the series for $\ln(0.7)$?
For Taylor’s theorem applied to $\ln(1+x)$ on $[-0.3,0]$, what lower bound can be used for $1+x$?
For $y=5$, find $z=(y-1)/(y+1)$. Is the transformed logarithmic tail bound applicable?
In the transformed-series estimate, why is replacing each tail denominator by $2m+3$ an upper-bound step?

Error Bounds for Logarithmic Series

What You'll Learn

Why logarithmic error bounds need attention

The logarithmic series and its interval

Why the logarithmic remainder has this bound

Worked Example: Approximating \(\ln(1.2)\)

Worked Example: Approximating \(\ln(1.2)\)

Negative inputs: use the interval, not the alternating rule

Worked Example: Bounding the error for \(\ln(0.8)\)

Worked Example: Bounding the error for \(\ln(0.8)\)

A transformed series for logarithms

Worked Example: Bounding an approximation to \(\ln 3\)

Worked Example: Bounding an approximation to \(\ln 3\)

Common mistake: treating every next term as an error bound

Check Your Understanding

Test yourself: AP Calculus practice exams

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Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Error Bounds for Logarithmic Series

What You'll Learn

Why logarithmic error bounds need attention

The logarithmic series and its interval

Why the logarithmic remainder has this bound

Worked Example: Approximating \(\ln(1.2)\)

Worked Example: Approximating \(\ln(1.2)\)

Negative inputs: use the interval, not the alternating rule

Worked Example: Bounding the error for \(\ln(0.8)\)

Worked Example: Bounding the error for \(\ln(0.8)\)

A transformed series for logarithms

Worked Example: Bounding an approximation to \(\ln 3\)

Worked Example: Bounding an approximation to \(\ln 3\)

Common mistake: treating every next term as an error bound

Check Your Understanding

Test yourself: AP Calculus practice exams

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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