About Harmonic Numbers and a sequence in OEIS I'd been looking for a series in OEIS and the one that fits better (resembles, at least) is one in this link in Examples: http://oeis.org/A082687. There, as we can see, it is equal to 
$$H'(2n) = H(2n)-H(n) = \cdots  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$
Also, I know that $H$ represents harmonic numbers. However here is the thing: I'm confused, since the harmonic numbers should be only numbers not a series. Then I searched at Wiki about these "harmonic series", but they seem not to exist. Bellow (1), there is something about Hilbert Matrix, maybe that is it, but I'm not sure as I do not even know what this is.
Could someone give me a clue about what these $H$s are and/or tell what (1) means?
Thanks
 A: Here is how I would write the expression you give in (1):
$$\overline H_{2n} = H_{2n} - H_n. \tag2$$
Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$ while $\overline H_n$ denotes the $n$th skew-harmonic number $\sum_{k = 1}^n (-1)^{k + 1}/k$.
In proving (2), we have
\begin{align}
\overline H_{2n} &= \sum_{k = 1}^{2n} \frac{(-1)^{k + 1}}{k}\\
&= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n - 1} - \frac{1}{2n}\\
&= 1 + \left (\frac{1}{2} - 1 \right ) + \frac{1}{3} + \left (\frac{1}{4} - \frac{1}{2} \right ) + \frac{1}{5} + \left (\frac{1}{6} - \frac{1}{3} \right ) + \cdots\\
&\qquad \cdots + \frac{1}{2n - 1} + \left (\frac{1}{2n} - \frac{1}{n} \right )\\
&= \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2n} \right ) - \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \right )\\
&= \sum_{k = 1}^{2n} \frac{1}{k} - \sum_{k = 1}^n \frac{1}{k}\\
&= H_{2n} - H_n,
\end{align}
as required to prove.
Checking that we indeed get the first few terms in OEIS A082687, since
\begin{align}
\overline H_2 &= \sum_{k = 1}^2 \frac{(-1)^{k + 1}}{k} = \frac{1}{2}\\
\overline H_4 &= \sum_{k = 1}^4 \frac{(-1)^{k + 1}}{k} = \frac{7}{12}\\
\overline H_6 &= \sum_{k = 1}^6 \frac{(-1)^{k + 1}}{k} = \frac{37}{60}\\
\overline H_8 &= \sum_{k = 1}^8 \frac{(-1)^{k + 1}}{k} = \frac{533}{840}
\end{align}
and selecting the numerators for each of these quantities we have the sequence: $\{1, 7, 37, 533, \ldots\}$ as desired.
A: $$\overline{H}_{2n}=\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}=\sum_{k=1}^{2n}(-1)^{k-1}\int_0^1 x^{k-1}\ dx=\int_0^1\sum_{k=1}^{2n}(-x)^{k-1}\ dx$$
$$=\int_0^1\frac{1-x^{2n}}{1+x}\ dx=\ln2-\int_0^1\frac{x^{2n}}{1+x}\ dx=H_{2n}-H_n$$
where the last result follows from $\int_0^1\frac{x^{2n}}{1+x}dx=\ln2+H_n-H_{2n}$.
