<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] $$H'_n = \sum_{k = 1}^n \frac {(-1)^n}{k}$$

*

*Harmonic Function : ($4$)
$$\color{green}{S} =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+....\frac{1}{99\cdot100}=  \color{green}{H'_{100}} $$
$$\begin{align*}T
&= \frac{1}{51\cdot100}+\frac{1}{52\cdot99}+\frac{1}{53\cdot98}+....+\frac{1}{100\cdot51}\\
& = \frac 1{151} \left[\frac 1{51} + \frac 1{100} +\frac 1{52} +\frac 1{99}....... \frac {1}{99}+\frac 1{52} + \frac1{100} + \frac 1{51}\right]\\
& = \frac 2{151}\left(\frac 1{51} + \frac 1{52} +.....\frac 1{100}\right)\\
& = \frac 2{151}(H_{100}-H_{50})\\
& = \frac 2{151}H'_{100} \\
\end{align*}$$
$$\color{blue}{T = \frac 2{151}S \implies \frac S{T} = \frac {151}{2}}$$
$H'_{2n} = H_{2n} -H_{n}$
</s>