# Series of negative terms of the alternating harmonic series

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.

• Oh yes they are! – Lord Shark the Unknown Jul 25 '17 at 4:50
• I believe you're only confused about what's the original harmonic series. The original harmonic series is the one with no alternating signs, just positive numbers being added. – Sergio Enrique Yarza Acuña Jul 25 '17 at 4:55
• @SergioEnriqueYarzaAcuña yes! I don't understand why he brought up the series with alternating signs before talking about the series with only the negative terms. Thank you a lot. – A Slow Learner Jul 25 '17 at 5:05

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
• @Geophysics No problem. You are probably confusing the fact that the author first mentions the series for $\ln(2)$ above as an example that Conditionally Convergent Series cannot be rearranged (see here for an explanation of how we can reach any real number by rearranging a conditionally convergent series, starting with your example!) – Brevan Ellefsen Jul 25 '17 at 5:12
They are $-\frac 12$ times the terms of the positive harmonic series $\sum_{n=1}^\infty\frac 1n$. You can compare them term by term.