How can I show these series combinations are the same algebraically? We have to show:
$$\frac{1}{2}\sum_{r=1}^{n} {\frac{1}{r}} \ + \ \sum_{r=1}^{n} {\frac{1}{2r+1}} \ = \ \sum_{s=1}^{2n}{\frac{1}{s+1}}$$
I understand how this works intuitively and the most formal way I can think to show it is as follows:
$$\frac{1}{2}\sum_{r=1}^{n} {\frac{1}{r}} \ + \ \sum_{r=1}^{n} {\frac{1}{2r+1}} \ = \ \sum_{r=1}^{n}(\frac{1}{2r}+\frac{1}{2r+1}) \ = \ \frac{1}{2} + \frac{1}{3} + \ ... \ + \frac{1}{2n} + \frac{1}{2n+1} \ = \ \sum_{s=1}^{2n}{\frac{1}{s+1}}$$
This is fine but it feels like there might be a more rigorous method without listing terms. I've tried to combine the two fractions on the left-hand side and change the summation limits but the result looks rather different to the right-hand side.
How can I show this without listing terms?
 A: You can use the following trick to only count the even terms:
$$\frac{1}{2}\sum_{r=1}^n\frac{1}{r} = \sum_{r=1}^n\frac{1}{2r} = \sum_{r=1}^{2n}\frac{1}{r} \cdot \frac{(-1)^{r}+1}{2}$$
And a similar trick to only count the odd terms. Notice that I've manually added two terms $\left[\frac{1}{2n+1}-\frac{1}{1}\right]$ to make it work out on the given range.
$$\sum_{r=1}^n\frac{1}{2r+1} = \left[\frac{1}{2n+1}-\frac{1}{1}\right] + \sum_{r=1}^{2n}\frac{1}{r} \cdot \frac{(-1)^{r+1}+1}{2}$$
Combing the two sums gives:
$$\left[\frac{1}{2}\sum_{r=1}^n\frac{1}{r}\right] + \left[\sum_{r=1}^n\frac{1}{2r+1}\right]$$
$$=$$
$$\left[\sum_{r=1}^{2n}\frac{1}{r} \cdot \frac{(-1)^{r}+1}{2}\right] + \left[\left[\frac{1}{2n+1}-\frac{1}{1}\right] + \sum_{r=1}^{2n}\frac{1}{r} \cdot \frac{(-1)^{r+1}+1}{2}\right]$$
$$=$$
$$\left[\frac{1}{2n+1}-\frac{1}{1}\right] + \sum_{r=1}^{2n}\frac{1}{r} \left( \frac{(-1)^{r}+1}{2} + \frac{(-1)^{r+1}+1}{2}\right)$$
$$=$$
$$\left[\frac{1}{2n+1}-\frac{1}{1}\right] + \sum_{r=1}^{2n}\frac{1}{r}$$
$$=$$
$$\sum_{r=2}^{2n+1}\frac{1}{r}$$
$$=$$
$$\sum_{r=1}^{2n}\frac{1}{r+1}$$
But honestly, I think your notation (listing the terms) is formal enough and easier to understand :D
