Shifting the summation index?

1. Aug 8, 2009

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.

Last edited: Aug 8, 2009
2. Aug 8, 2009

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.

3. Aug 8, 2009

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.

Last edited: Aug 8, 2009
4. Aug 8, 2009

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*} \sum_{k=2}^n \frac{1}{k^2} &< \sum_{k=2}^n \frac{1}{k(k-1)} \\ &= \sum_{k=1}^{n-1} \frac{1}{k(k+1)} \\ &= 1 - \frac{1}{n} \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 $1/k(k+1) = 1/k - 1/(k+1)$.