Shifting the Summation Index in Zeta Function Convergence Proof?

[Skynt]

Can anyone explain this property of shifting the index on the summation notation?

I'm reading a book and came across this which has confused me. I don't see how these are equal:
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{1}{2} + \sum_{k=2}^{n+1} \frac{1}{k(k-1)}$$

It's part of an explanation that proves that the zeta function converges for values equal to or larger than 2. I just fail to see how they're equal.

 

[rasmhop]

They're not equal. Just consider the case n=1 in which case the identity states:
$$\frac{1}{1(1+1)} = \frac{1}{2} + \frac{1}{2(2-1)}$$
which is clearly false since the left hand side is 1/2 while the right hand side is 1/2 + 1/2 = 1. We do however have:
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=2}^{n+1}\frac{1}{k(k-1)}$$

Perhaps you misread/miscopied your book or it's an error in the book.

 

[Skynt]

Yeah that's exactly what confused me. I tested it too. It's definitely an error or I misread it. If anyone has the Art and Craft of Problem Solving, it's on page 162 of the Algebra chapter. Perhaps someone could clear me up on this proof lol.

 

[rasmhop]

It seems to be an error in the book (and not just a typo as his later derivations depends on the formula). Anyway doing the correct shifting we can show a slightly stronger result:
$$\begin{align*}<br /> \sum_{k=2}^n \frac{1}{k^2} &< \sum_{k=2}^n \frac{1}{k(k-1)} \\<br /> &= \sum_{k=1}^{n-1} \frac{1}{k(k+1)} \\<br /> &= 1 - \frac{1}{n}<br /> \end{align*}$$
Here we use the inequality $1/k^2 < 1/k(k-1)$ mentioned in the text, we shift the index by one and finally we either use the formula:
$$\sum_{k=1}^{n} \frac{1}{k(k+1)} = 1 - \frac{1}{n+1}$$
given in the text or simply notice that the sum telescopes as $1/k(k+1) = 1/k - 1/(k+1)$.