Tribute to my first problem

We form the summations $\displaystyle\sum_{k=51}^{200} \dfrac{1}{k} = \sum_{k=51}^{100} \dfrac{1}{k} + \sum_{k=101}^{150} \dfrac{1}{k} + \sum_{k=151}^{200} \dfrac{1}{k} \gt \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} = \dfrac{6 + 4 + 3}{12} = \dfrac{13}{12} \gt 1$ .

So $\begin{aligned}\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\cdots\frac{1}{200}>1\end{aligned}$ is greater than $\boxed{\text{1}}$

$\begin{aligned}\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\cdots+\frac{1}{200}\end{aligned}$

Note that $\begin{aligned} & n < n+1< n+2 < \cdots \ n+k \\& \implies \frac{1}{n} > \frac{1}{n+1}+\frac{1}{n+2} > \cdots > \frac{1}{n+k}\end{aligned}$ for $n\mathbb\in N$ and $k=1,2,3,\cdots$ $\begin{aligned} S & = \frac{1}{51}+\frac{1}{52}+\frac{1}{53} +\cdots+\frac{1}{200} \\& = \left(\frac{1}{51} > \frac{1}{100}+\frac{1}{52}> \frac{1}{100}+\frac{1}{53} > \frac{1}{100} \cdots +\frac{1}{99}> \frac{1}{100}+\frac{1}{100}\right)+\left(\frac{1}{101}> \frac{1}{200}+\frac{1}{102}> \frac{1}{200}+\frac{1}{103}>\frac{1}{200} +\cdots+\frac{1}{200}\right) \\&= \displaystyle\sum_{n=51}^{100}\frac{1}{n}>50\times \frac{1}{100} + \displaystyle\sum_{k=101}^{200}\frac{1}{k} > 100\times\frac{1}{200} \\& > \frac{1}{2}+\frac{1}{2} > 1 \end{aligned}$ Thus the answer is $\text{yes}$ .

Noting that $\displaystyle\sum_{k=51}^{100} \dfrac{1}{k} \gt 50 \times \dfrac{1}{100} = \dfrac{1}{2}, \sum_{k=101}^{150} \dfrac{1}{k} \gt 50 \times \dfrac{1}{150} = \dfrac{1}{3}$ and $\displaystyle\sum_{k=151}^{200} \dfrac{1}{k} \gt 50 \times \dfrac{1}{200} = \dfrac{1}{4}$ ,

we find that $\displaystyle\sum_{k=51}^{200} \dfrac{1}{k} = \sum_{k=51}^{100} \dfrac{1}{k} + \sum_{k=101}^{150} \dfrac{1}{k} + \sum_{k=151}^{200} \dfrac{1}{k} \gt \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} = \dfrac{6 + 4 + 3}{12} = \dfrac{13}{12} \gt 1$ . The answer is therefore $\boxed{\text{yes}}$ .