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} $

Question

I have this question:

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

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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?

Answer 1

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 : $$\sum_{n=1}^{2008} \frac{1}{n^2+n} = \sum_{n=1}^{2008} \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{2009}$$.

Answer 2

Hint: $\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$. Now telescope.

You make want to look up partial fraction decomposition, this can be very useful in contest problems, it helps to solve some recurrences via generating functions, and it also helps to find telescoping sums.