TITLE: Series of negative terms of the alternating harmonic series
QUESTION [0 upvotes]: In Stephen Abbott's Understanding Analysis textbook, section 2.9, he wrote that the negative terms of the harmonic series $$\sum_{n=1}^\infty\frac{-1^{n+1}}{n} = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots$$
take alone form the series $$\sum_{n=1}^\infty\frac{-1}{2n}=-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\cdots$$
He said (quoted): "The partial sums of this series are precisely -1/2 the partial sums of the harmonic series, and so march off (at half speed) to negative infinity."
If you calculate the partial sums of the two series and compare them, the partial sums of the series with only negative terms are not -1/2 the partial sums of the original harmonic series. Thus, I am a little confused about what he meant.

REPLY [1 votes]: First, note that the harmonic series is 
$$\color{green}{\sum_{k=1}^\infty \frac{1}{n}} = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$$
What you call "the harmonic series" in your post is not the same as this. In fact,
$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \ln(2)$ converges!  

Next, note that 
$$\sum_{n=1}^\infty\frac{\color{red}{-1}}{\color{red}{2}n} = \color{red}{\frac{-1}{2}}\color{green}{\sum_{n=1}^\infty\frac{1}{n}}$$
Since the coefficient out front is negative (and all terms are positive) the series approaches $-\infty$. However, since each term is only half the size of the associated term in the Harmonic Series, it is somewhat sensical to say that this new series diverges half as fast as the Harmonic Series
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