Estimate \(\log 2\) by evaluating \[ S=\sum_{n=0}^5 a_n \qquad\text{and}\qquad E=\sum_{n=0}^5 b_n(1/2)^n, \] where \[ f(x)=\log(1+x)=\sum a_nx^n \] and \[ g(y)=f\!\left(\frac{y}{1-y}\right)=\sum b_ny^n. \] How do these approximations compare to \(\log 2\)?
What is the radius of convergence of the power series for \(g(y)\)?
Solution
1. The first five nonzero coefficients of \(f(x)=\log(1+x)\) are\[ a_n=\frac{(-1)^{n+1}}{n} \qquad(1\le n\le5), \]with \(a_0=0\). Hence\[ S=1-\frac12+\frac13-\frac14+\frac15 =\frac{47}{60} \approx0.783333. \]
For \(|x|<1\), \[ \log(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}x^n. \] Substituting \(x=1\) in the first five terms gives the displayed value of \(S\).
2. The transformed function is\[ g(y)=-\log(1-y) =\sum_{n=1}^{\infty}\frac{y^n}{n}, \]so \(b_0=0\) and \(b_n=1/n\) for \(n\ge1\). Therefore\[ E =\frac12+\frac18+\frac1{24}+\frac1{64}+\frac1{160} =\frac{661}{960} \approx0.688542. \]
Directly, \[ g(y) =f\!\left(\frac{y}{1-y}\right) =\log\!\left(1+\frac{y}{1-y}\right) =\log\!\left(\frac1{1-y}\right) =-\log(1-y). \] Its Taylor expansion at \(0\) is therefore \(\sum_{n\ge1}y^n/n\). Evaluating its degree-five truncation at \(y=1/2\) gives the displayed value of \(E\).
3. The transformed approximation is much more accurate:\[ S-\log2\approx0.090186, \qquad \log2-E\approx0.004606. \]In particular \(S\) is an overestimate, while \(E\) is an underestimate.
The alternating harmonic series gives \[ \log2=1-\frac12+\frac13-\frac14+\frac15-\cdots. \] Its odd partial sums decrease to \(\log2\), so \(S>\log2\). On the other hand, \[ \log2=\sum_{n=1}^{\infty}\frac1{n2^n}, \] whose terms are positive, so \(E<\log2\). Using \(\log2\approx0.693147\) gives the stated numerical errors.
4. The radius of convergence of the power series for \(g\) is\[ \boxed{1}. \]
Since \(b_n=1/n\), \[ \limsup_{n\to\infty}|b_n|^{1/n} =\lim_{n\to\infty}n^{-1/n} =1. \] The Cauchy-Hadamard formula therefore gives radius \(R=1\). Equivalently, \(g(y)=-\log(1-y)\) has its nearest singularity to \(0\) at \(y=1\).