Exercise 1.3.39* (It's easier to prove the general result.)

Proof. We prove by induction tor $n\in {\mathbb{N}}^{\text{∗}}$ the property

• If $n=1$, then

  $$\begin{array}{llll}\hfill \underset{k=1}{\overset{2n}{\sum }}\frac{{(-1)}^{k+1}}{k}& =\underset{k=1}{\overset{2}{\sum }}\frac{{(-1)}^{k+1}}{k}=1-\frac{1}{2}=\frac{1}{2}& \hfill & \end{array}$$

  and

  $$\underset{j=n+1}{\overset{2n}{\sum }}\frac{1}{j}=\underset{j=2}{\overset{2}{\sum }}\frac{1}{j}=\frac{1}{2},$$

  thus $𝒫(1)$ is true.

• Assume now that $𝒫(n)$ is true for some integer $n\ge 1$. Then

  $$\begin{array}{llll}\hfill \underset{k=1}{\overset{2n+2}{\sum }}\frac{{(-1)}^{k+1}}{k}& =\underset{k=1}{\overset{2n}{\sum }}\frac{{(-1)}^{k+1}}{k}+\frac{1}{2n+1}-\frac{1}{2n+2}& \hfill & \\ \hfill & =\underset{j=n+1}{\overset{2n}{\sum }}\frac{1}{j}+\frac{1}{2n+1}-\frac{1}{2n+2}\text{(by the induction hypothesis)}& \hfill & \\ \hfill & =\underset{j=n+2}{\overset{2n+2}{\sum }}\frac{1}{j}+\frac{1}{n+1}-\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+1}-\frac{1}{2n+2}& \hfill & \\ \hfill & =\underset{j=n+2}{\overset{2n+2}{\sum }}\frac{1}{j}+\frac{1}{n+1}-\frac{2}{2(n+1)}& \hfill & \\ \hfill & =\underset{j=n+2}{\overset{2n+2}{\sum }}\frac{1}{j}.& \hfill & \end{array}$$

  So $𝒫(n)\Rightarrow 𝒫(n+1)$ for all $n\in {\mathbb{N}}^{\text{∗}}$.

In particular, $𝒫(1000)$ is true, so

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