<s>[INST] Two quite messy sequences Consider the following $50-$term sums$:$
$$ S=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+....\frac{1}{99\cdot100}$$ and
$$T=\frac{1}{51\cdot100}+\frac{1}{52\cdot99}+\frac{1}{53\cdot98}+....+\frac{1}{100\cdot51}$$
Express $\frac{S}{T}$ as an irreducible fraction.
My attempt$:$
The first equation can be written as $$S=\frac11-\frac12+\frac13-\frac14+\frac15-\frac16+....+\frac1{99}-\frac1{100}$$ $\implies$ $$S=(1+\frac13+\frac15+....+\frac1{99})-(\frac12+\frac14+\frac16+....+\frac1{100})$$ or $$S=1+\frac12\operatorname{ln}50-\frac12\operatorname{ln}\frac{103}{2}$$ and $$T=2(\frac{1}{51\cdot100}+\frac{1}{52\cdot99}+\frac{1}{53\cdot98}+....+\frac{1}{75\cdot76})$$ After this I am not able to do anything. And also the value of S which I get doesn't seems to be correct(though I have calculated it using formula)as it is quite messy and I don't think S can be further reduced. Any help will be greatly appreciated.
 [/INST] For every integer $n\geq 1$, let $$H_n = \sum_{i=1}^{n} \frac{1}{i}$$
On one hand, one has
$$S=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+....\frac{1}{99\cdot100} = \sum_{i=1}^{50} \frac{1}{(2i-1)(2i)} = \sum_{i=1}^{50} \frac{1}{2i-1}-\frac{1}{2i}$$
so
$$S=\left(\sum_ {i=1}^{100} \frac{1}{i} - \sum_{i=1}^{50} \frac{1}{2i} \right)-\sum_{i=1}^{50} \frac{1}{2i}  = H_{100}-\frac{1}{2}H_{50}-\frac{1}{2}H_{50}$$ so $$\boxed{S= H_{100}-H_{50}}$$
On the other hand,
$$T=\frac{1}{51\cdot100}+\frac{1}{52\cdot99}+\frac{1}{53\cdot98}+....+\frac{1}{100\cdot51} =\sum_{i=1}^{50} \frac{1}{(50+i)(101-i)} = \frac{1}{151}\left(\sum_{i=1}^{50}\frac{1}{50+i} + \frac{1}{101-i}\right)$$
so
$$T=\frac{1}{151}\left(H_{100}-H_{50} + \sum_{i=1}^{50}\frac{1}{101-i}\right) = \frac{1}{151}\left(H_{100}-H_{50} + \sum_{i=51}^{100}\frac{1}{i}\right) = \frac{1}{151}\left(H_{100}-H_{50} + H_{100}-H_{50}\right)$$ so $$\boxed{T = \frac{2}{151} \left(H_{100}-H_{50} \right)}$$
Using these two results, you see finally that
$$\frac{S}{T} = \frac{151}{2}$$
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