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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).

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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.

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    $\begingroup$ 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. $\endgroup$ Commented Jul 10, 2016 at 18:19
  • $\begingroup$ @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 $N\to\infty$. $\endgroup$ Commented Jul 10, 2016 at 18:44
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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).

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