Taylor Series for Logarithms About Positive Centers

Introduction

A Taylor series for \(\ln x\) need not be centered at \(1\). Choosing any center \(a>0\) gives a local power series in the displacement \(x-a\). The center determines the coefficients and, importantly, how far the series can extend before it reaches the singularity at \(x=0\).

The construction has a simple form: write \(x=a+h\), factor out \(a\), and use the series for \(\ln(1+u)\). This gives both a useful expansion and a clear way to determine its convergence.

Constructing the Series

For \(x>0\) and \(a>0\), factor \(x\) as \(a\) times a dimensionless quantity:

Taking logarithms gives

The known logarithm series is

Substitute \(u=(x-a)/a\). The resulting Taylor series is

The coefficients can also be checked directly from derivatives. For each integer \(n\geq 1\),

Thus the Taylor coefficient of degree \(n\) at \(a\) is the \(n\)th derivative evaluated at \(a\), divided by \(n!\):

Since \((-1)^{n-1}=(-1)^{n+1}\), this agrees with the coefficient in the series above.

Why the Logarithm Series Works

The construction relies on the geometric series. For \(|t|<1\), the finite geometric-sum identity gives

Integrate from \(0\) to \(u\), where \(|u|<1\). The integral on the left is \(\ln(1+u)\). The finite sum integrates term by term, and the remaining integral is an error term:

Set \(r=|u|<1\). Along the interval of integration, \(|t|\leq r\), so \(|1+t|\geq 1-r\). Therefore the absolute value of the remainder is at most

Because \(r<1\), this bound tends to zero as \(N\) increases. The finite sums therefore approach \(\ln(1+u)\), proving the series for \(|u|<1\). Substituting \(u=(x-a)/a\) produces the expansion about \(a\).

Radius and Interval of Convergence

In the power series in \(x-a\), the coefficient of degree \(n\) is \((-1)^{n+1}/(n a^n)\). The ratio of the absolute values of consecutive terms, for a fixed displacement \(h=x-a\), tends to \(|h|/a\). Thus the series converges when \(|h|<a\) and diverges when \(|h|>a\). Its radius of convergence is \(a\).

This radius has a geometric explanation. The center \(a\) is a distance \(a\) from \(x=0\), where \(\ln x\) is singular. The series can represent the logarithm within that distance, but its behavior at the two boundary points must be tested separately. The corresponding real interval, after checking the endpoints, is \((0,2a]\).

At the right endpoint \(x=2a\), the ratio \((x-a)/a\) equals \(1\), giving

The alternating harmonic series converges, but not absolutely, since the sum of the absolute values is the divergent harmonic series. At the left endpoint \(x=0\), the ratio is \(-1\), so the terms become \(-1/n\). The series diverges, consistently with the fact that \(\ln 0\) is not defined and \(\ln x\) tends to negative infinity as \(x\) approaches zero from the right.

Worked Examples

Termwise Differentiation as a Check

Inside its radius of convergence, a power series can be differentiated term by term. Applying that rule to the logarithm series gives

The right side is a geometric series. For \(|x-a|<a\), its ratio has absolute value less than \(1\), so

This is the derivative of \(\ln x\). The series also has the correct value at \(x=a\), since all nonconstant terms vanish there and the result is \(\ln a\). These checks support the claimed expansion throughout its open interval of convergence.

A Common Endpoint Mistake

A frequent error is to take the radius \(a\) and report the interval \((0,2a)\), stopping before checking either endpoint. The radius test is decisive only for points strictly inside or strictly outside the radius; it does not decide convergence on the boundary.

For example, with \(a=1\), the right endpoint \(x=2\) gives the alternating harmonic series \(\sum_{n=1}^{\infty}(-1)^{n+1}/n\), which converges. Declaring that endpoint excluded just because \(|x-1|=1\) is incorrect. The correction is to substitute each endpoint into the series: \(x=2\) converges, whereas \(x=0\) produces \(-\sum_{n=1}^{\infty}1/n\) and diverges. The correct interval is \((0,2]\).

Why the Center Matters

The center controls both the scale of the expansion and the interval on which it represents \(\ln x\). A larger positive center moves the singularity at zero farther away and increases the radius in absolute \(x\)-units. But the relative condition remains \(|x-a|<a\), so the interval extends from zero to twice the center, with convergence at the right endpoint.

This makes the expansion useful for local approximation as well as for understanding convergence. To use it at a target value \(x\), choose a center \(a\) for which \(|x-a|/a\) is comfortably below \(1\). Then the powers decrease, and the series may provide an effective approximation. Near the boundary, convergence can be much slower; at the left boundary, the logarithm’s singularity prevents convergence altogether.