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In Concrete Mathematics (Graham, Knuth, Patashnik), on page 58, there is the below example of calculating the value of an infinite sum :

$$ \begin{align} \sum_{k \geq 0} \frac{1}{(k+1)(k+2)} &= \sum_{k \geq 0} k^{\underline{-2}} \\ &= \underset{n \rightarrow \infty}{\lim} \sum_{k=0}^{n} k^{-2} = \underset{n \rightarrow \infty}{\lim} \frac{k^\underline{-1}}{-1}\Biggr|_0^n = 1 \end{align} $$

For that last part I don't understand how it is $1$ and not $-1$. To start with, at $0$, the summation property gives us 0:

$$ \frac{0^{\underline{-1}}}{-1} = \frac{\frac{0}{0+1}}{-1} = 0 $$

Then $1$ is

$$ \frac{1^{\underline{-1}}}{-1} = \frac{\frac{1}{1+1}}{-1} = -\frac12 $$

and $2$ is $$ \frac{2^{\underline{-1}}}{-1} = \frac{\frac{1}{2+1}}{-1} = -\frac13 $$

And so on tending towards -1 for larger $n$. As I understand it (the subtraction vertical bar notation) if $n$ is, say, $2$ (a long way from infinity to be sure) then we'd get $-\frac13 - 0 = -\frac13$. And so on getting closer to $-1$ as best I can tell.

Friedrich 'Fred' Clausen
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You're missing a negative sign. The vertical bar notation $$f(k)\;\Biggr|_0^n$$ means: evaluate $f(k)$ at $k=n$ and $k=0$, and subtract the two, obtaining $$f(n) - f(0).$$ Using this convention, you get $$\frac{k^\underline{-1}}{-1}\;\Biggr|_0^n=\frac{n^\underline{-1}}{-1}-\frac{0^\underline{-1}}{-1}=\frac{-1}{n+1}-(-1).$$ The first term tends to $0$ as $n\to\infty$. As for the second term, the two negative signs combine to give $+1$.

grand_chat
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