Upper bound on alternating harmonic series

Question

Let $a > 0$. Is the following statement true: \begin{align} \ln(2) = \sum_{n=1}^\infty{\frac{(-1)^{n+1}}{n}} > \sum_{n=1}^\infty{\frac{(-1)^{n+1}}{n+a}} \end{align} From my intuition the statement is true, because in absolute terms the numbers on the right hand side are smaller

Answer 1

The argument based on absolute values may not be correct since the terms may alternatively go below or above each other. A safer way naturally is to remove the oscillation in the terms: $$x_n{=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n}\\=\sum_{k=0}^\infty {1\over (2k+1)(2k+2)}}$$ $$y_n{=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+a}\\=\sum_{k=0}^\infty {1\over (2k+1+a)(2k+2+a)}}$$since for $a>0$ we can say $${1\over (2k+1+a)(2k+2+a)}<{1\over (2k+1)(2k+2)}$$by term by term summation we can conclude what we want.

Answer 2

We have

$$\sum_{n=1}^\infty \left( \frac{(-1)^{n+1}}{n} - \frac{(-1)^{n+1}}{n+a} \right) = \sum_{n=1}^\infty (-1)^{n+1} \left( \frac{1}{n} - \frac{1}{n+a} \right) = a \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+a)}$$

This is greater than zero because this is an alternating series, the first term is positive, and $\frac{1}{n(n+a)}$ is strictly decreasing.