# Find the value of $\frac {1} {1^2 + 1} +\frac {1} {2^2 + 2} +\frac {1} {3^2 + 3} +\frac {1} {4^2 + 4} ++ \dots + \frac {1} {2008^2 + 2008}$

I have this question:

Find the value of: $$\frac {1} {1^2 + 1} +\frac {1} {2^2 + 2} +\frac {1} {3^2 + 3} +\frac {1} {4^2 + 4} ++ \dots + \frac {1} {2008^2 + 2008}$$

My attempt:

I tried to think of a better way to handle: $$\frac {1} {n^2 + n}$$ Then I got (doesn't work only on $\frac{1} {1^2 +1}$): $$\frac {n-1} {n^3 -n}$$

By putting the values in, I got: $$\frac {1} {2} + \frac {2-1} {2^3 - 2} + \frac {3-1} {3^3 - 3} + \frac {4-1} {4^3 - 4} + \dots + \frac {2008-1} {2008^3 - 2008}$$ It's still doesn't make sense. Is there another way of solving this question? Can I have a hint or a guide?

• What is the source of the problem, and its motivation? These are more interesting for many readers than your attempt. Because this is only a finite sum, you could simply add it using a computer, after all - why is it interesting to do in some other way? – Carl Mummert Feb 1 '17 at 11:49
• @CarlMummert Sorry to ask, but what do mean by 'These are more interesting for many readers than your attempt' :) – Adola Feb 1 '17 at 12:02
• I mean that the bests posts on this site include discussion of the source and motivation of the problem, which could be included in addition to the statement and any attempt. The attempt does show what you have tried, but in cases like this where the attempt doesn't make much progress, it may be better to leave out the attempt and instead focus on the source and motivation. – Carl Mummert Feb 1 '17 at 12:04
• @CarlMummert On your first post, I was trying to solve this without a computer. Using a computer to solve this would defeats the point of asking a finite question on this site. For your second post, I would like the readers here to know about 'rationalizing the denominator'. it could help them for solving another different problem like this ($\frac1{\sqrt{n}+\sqrt{n+1}} = \sqrt{n+1}-\sqrt{n}$) and more! – Adola Feb 18 '17 at 2:45

Write $\frac{1}{n^2+n} = \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. See if now you can identify a telescoping sequence as follows : $$\sum_{n=1}^{2008} \frac{1}{n^2+n} = \sum_{n=1}^{2008} \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{2009}$$.
Hint: $\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$. Now telescope.