Summation - relatively simple? I have a question which might be too simple for this site but I really tried many ideas without coming to a solution. This is assignment from elementary school in which I am trying to help and the solution should be relatively simple but somehow I cannot figure out the correct approach.
The assignment is as follows i.e. calculate the sum:
$$\frac{1}{1+2} + \frac{1}{2+3} + \frac{1}{3+4} + … + \frac{1}{98+99}+\frac{1}{99+100}$$

I think there isn't a nice closed form for this sum. But if we let $H_n$ denote $1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$, then
$\frac{1}{1 + 2} + \frac{1}{2 + 3} + ... + \frac{1}{99 + 100} = \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{199} = \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{200} - 1 - (\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{200}) = H_{200} - 1 - \frac{1}{2}(1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{100}) = H_{200} - \frac{1}{2}H_{100} - 1$
But the harmonic sum does not have a nice closed form as far as I knew.
