Prove $\sum_{k=1}^r\frac{1}{k+r}=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}$ How to prove that 

$$\sum_{k=1}^r\frac{1}{k+r}=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}\tag1$$

We know that both sides are equal to $H_{2r}-H_r$ but I am trying to convert the left side to the right side using only series manipulations and without going through $H_{2r}-H_r$
There is nothing I could try but we know that
$$\sum_{k=1}^r\frac{1}{k+r}=\sum_{k=1+r}^{2r}\frac{1}{k}$$
What next? Thank you.

By the way, its easy to prove it using integration,
$$\sum_{k=1}^r\frac{1}{k+r}=\int_0^1\sum_{k=1}^r x^{k+r-1}\ dx=\int_0^1\frac{x^r-x^{2r}}{1-x}\ dx$$
$$=\int_0^1\frac{1-x^{2n}-(1-x^r)}{1-x}\ dx=H_{2r}-H_r\tag2$$
and we proved here
$$\overline{H}_{2r}=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}=H_{2r}-H_r\tag3$$
Hence by $(2)$ and $(3)$ , $(1)$ is proved

As has been pointed out by @Greg Martin, one way to get the result you desire using only manipulations of series would be to essentially run the proof of the result I gave here in reverse. 
Starting with $\sum_{k = 1}^r \frac{1}{k + r}$ reindexing $k \mapsto k - r$ we have
\begin{align}
\sum_{k = 1}^r \frac{1}{k + r} &= \sum_{k = r + 1}^{2r} \frac{1}{k}\\
&= \sum_{k = 1}^{2r} \frac{1}{k} - \sum_{k = 1}^n \frac{1}{k}\\
&= \left (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2r - 1} + \frac{1}{2r} \right ) - \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{r} \right )\\
&= 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 )\\ 
&= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n - 1} - \frac{1}{2n}\\
&= \sum_{k = 1}^{2r} \frac{(-1)^{k + 1}}{k},
\end{align}
as desired.
Of course this is just $H_{2r}$ and $H_r$ in disguise. So let us try again starting from the other side.
\begin{align}
\sum_{k = 1}^{2r} \frac{(-1)^{k + 1}}{k} &= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2r - 1} - \frac{1}{2r}\\
&= \left (1 - \frac{1}{2} \right ) + \left (\frac{1}{3} - \frac{1}{4} \right ) + \cdots + \left (\frac{1}{2r - 1} - \frac{1}{2r} \right )\\
&= \left (1 + \frac{1}{2} - 2 \cdot \frac{1}{2} \right ) + \left (\frac{1}{2} + \frac{1}{4} - 2 \cdot \frac{1}{4} \right ) + \cdots\\
& \qquad \cdots + \left (\frac{1}{2r - 1} + \frac{1}{2r} - 2 \cdot \frac{1}{2r} \right )\\
&= \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2r - 1} + \frac{1}{2r} \right ) - 2 \left (\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2r} \right )\\
&= \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2r - 1} + \frac{1}{2r} \right ) - \left (1 + \frac{1}{2} + \cdots + \frac{1}{r} \right )\\
&= \frac{1}{r + 1} + \frac{1}{r + 2} + \cdots + \frac{1}{2r}\\
&= \sum_{k = 1}^r \frac{1}{k + r},
\end{align}
which is essentially the same thing. 
Finally, this identity is known as the Botez-Catalan identity. 
