Sum of a series (combining divergent ones)

Question

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).

Answer 1

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.

Answer 2

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).