Relating alternating harmonic series with an example

I'm working through a problem, and I was given this relation:

$$A_{2n} = h_{2n} - h_n$$ where $$A_n = \sum_{n=1}^n {(-1)^{j+1} \over j}$$ $$h_n = \sum_{n=1}^n {1 \over j}$$

Perhaps I'm just not understanding the notation since when I work out an example, it doesn't compute.

If $n=3$ $$A_n = A_3 = { 1\over1 } - { 1\over2 } + { 1\over3 } $$ $$A_{2n} = A_6 = { 1\over1 } - { 1\over2 } + { 1\over3 }- { 1\over4 } + { 1\over5 } - { 1\over6 }$$ $$h_n = h_3 = { 1\over1 } + { 1\over2 } + { 1\over3 }$$ $$h_{2n} = h_6 = { 1\over1 } + { 1\over2 } + { 1\over3 } + { 1\over4 } + { 1\over5 } + { 1\over6 }$$ So, if $A_{2n} = h_{2n} - h_n$, then with this example, we have: $$A_6 = h_6 - h_3$$ which is:

$$({ 1\over1 } - { 1\over2 } + { 1\over3 }- { 1\over4 } + { 1\over5 } - { 1\over6 }) = ({ 1\over1 } + { 1\over2 } + { 1\over3 } + { 1\over4 } + { 1\over5 } + { 1\over6 }) - ({ 1\over1 } + { 1\over2 } + { 1\over3 }) $$ $$ = ({ 1\over1 } - { 1\over1 }) + ({ 1\over2 } - { 1\over2 }) + ({ 1\over3 } - { 1\over3 }) + { 1\over4 } + { 1\over5 } + { 1\over6 }$$ $$ = { 1\over4 } + { 1\over5 } + { 1\over6 }$$

which appears to be false.

So, if the relation $A_{2n} = h_{2n} - h_n$ is true, how can it be shown through example?

Solution 1:

$h_6-h_3=h_6-2\cdot (1/2)h_3$ serves to subtract the even terms twice which makes the result the same as the alternating $A_6.$ [In the subtracted $(1/2)h_3$ multiply the $1/2$ through first making it the even terms of $h_6.$]

Another note: Because of the subsequent multiplying by $2$ one is subtracting the even terms of $h_6$ twice, resulting in changing their signs for appearance in $A_6.$

Solution 2:

Well, let's start off with formulating a relationship between $A$ and $h$ that does work.

We note that $$h_{2n} = \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{2n},$$ and this would turn into $A_{2n}$ if we could take all the even denominator terms and subtract twice their value; in other words, if

$$A_{2n} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \cdots - \frac{1}{2n}$$

then

$$h_{2n} - A_{2n} = 2 \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n} \right) = \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n} = h_n.$$

Therefore,

$$A_{2n} = h_{2n} - h_n.$$

But wait, that is the identity that was claimed! So what's going on? Well, let's look at your example. You selected $n = 3$ and then concluded that $$\frac{1}{1} - \frac{1}{2} + \cdots - \frac{1}{6} = \frac{1}{4} + \frac{1}{5} + \frac{1}{6}$$ is "obviously" false. But did you actually try calculating each side? They both end up being $\frac{37}{60}$. In fact, if you did this with any positive integer $n$, it does actually work!

For the astute reader, consider how $A_{2n} = h_{2n} - h_n$ is related to the identity $$\log 2 = \log 2n - \log n.$$