Review : Taylor Series — Question 7

Question 7 – Solution

Strategy. Change the evaluation point to shrink the geometric factor controlling the Taylor tail.

Step 1: Construct the odd-power series. Since $L'(z)=2/(1-z^2)=2\sum_{k\ge 0}z^{2k}$ and $L(0)=0$, $$\boxed{L(z)=2\sum_{k=0}^{\infty}\frac{z^{2k+1}}{2k+1},\quad |z|<1.}$$ The geometric series converges uniformly on any closed segment inside $(-1,1)$, so termwise integration is justified.

Step 2: Bound a positive tail. The equation $(1+z)/(1-z)=2$ gives $z=1/3$. With $A_N=2\sum_{k=0}^{N-1}z^{2k+1}/(2k+1)$ and $0<z<1$, $$\begin{aligned} 0<L(z)-A_N&=2\sum_{k=N}^{\infty}\frac{z^{2k+1}}{2k+1}\\ &<\frac{2z^{2N+1}}{(2N+1)(1-z^2)}=:B_N. \end{aligned}$$ Strictness follows because later denominators exceed $2N+1$.

Step 3: Count the accelerated terms. At $z=1/3$, $B_N=9/[4(2N+1)3^{2N+1}]$, strictly decreasing in $N$. Here $B_5=9/7794468>10^{-6}$ and $B_6=9/82904796<1.086\times 10^{-7}$. Thus six terms are the smallest number certified by this bound, and $$\boxed{A_6<\ln 2<A_6+\frac 9{82904796}},\qquad A_6\approx 0.6931470738.$$

Step 4: Make a fair certificate comparison. For the direct alternating sum with $N$ terms, the stated bound is $1/(N+1)$. To make that bound strictly less than $10^{-6}$ requires $N\ge 1{,}000{,}000$. The comparison is six versus one million terms under these particular certificates. It does not assert that a million terms are necessary for the true direct-series error, nor that a loose bound is an optimal algorithm. The transformed series profits from the factor $z^2=1/9$.