# Sum of a series (combining divergent ones)

I am reading in a book, without any explanation, the following identity (with $a<b$):

$$\sum_{k=0}^{\infty}\left(\frac{1}{k+a+1}-\frac{1}{k+b+1}\right)=\frac{1}{a+1}+\dots +\frac{1}{b}$$

Since I am unable to prove this, and could not find an answer in the archives of this website, I would appreciate any help or useful reference (the book definitely writes this, there might possibly be a typo but I don't think so).

I presume $a$ and $b$ are positive integers. In that case, consider your infinite sum: it contains positive summands $1\over a+1$, $1\over a+2$, $1\over a+3$, ... and so on, and negative summands $-{1\over b+1}$, $-{1\over b+2}$, $-{1\over b+3}$, ... and so on. Since $a<b$, then all positive summands between $1\over a+1$ and $1\over b$ survive, while the others cancel out, because they are matched by a negative summand.
• Convergence still needs to be addressed. If we just interleave the $\frac1{k+a+1}$ and $\frac{-1}{k+b+1}$ terms to begin with, then the resulting series is not absolutely convergent, so it is not immediate that we're allowed to rearrange terms to make the cancelling happen. Commented Jul 10, 2016 at 18:19
• @HenningMakholm You are right: to take care of convergence I think one can write down explicitly the partial sum, up to $k=N$, and then take the limit $N\to\infty$. Commented Jul 10, 2016 at 18:44
The series is adding and subtracting an infinite list of terms. Match them up and see that remains. With k=0 you get the first term $\frac1{a+1}$. Since b > a that term is safe. As k increases the positive terms will match earlier negative terms and cancel each other.
The initial second term $-\frac1{b+1}$ with k=0 vs $\frac1{k+a+1} = \frac1{(b-a) + a + 1} = \frac1{b+1}$ when k=(b-a).