Prove that $\frac{\frac{1}{11}+\frac{1}{12}+\dots+\frac{1}{200}}{\frac{1}{10*11}+\frac{1}{11*12}+\dots+\frac{1}{19*20}}>19$

Question

Prove that $$\dfrac{\dfrac{1}{11}+\dfrac{1}{12}+\dots+\dfrac{1}{200}}{\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+\dots+\dfrac{1}{19\cdot20}}>19$$

My attempt:The denominator is $\frac{1}{20}$ using telescopic series.

$\frac{1}{10*11}+\frac{1}{11*12}+\dots+\frac{1}{19*20}=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\dots-\frac{1}{20}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$

I think we could Prove that the top part of fraction is bigger than $1$ but I don't know how.

Answer 1

Your calculation for the denominator (telescopic sum and not series) was the hard part, since \begin{align}\dfrac{\dfrac{1}{11}+\dfrac{1}{12}+\dots+\dfrac{1}{200}}{\dfrac{1}{20}}&>\dfrac{\dfrac{1}{200}+\dfrac{1}{200}+\dots+\dfrac{1}{200}}{\dfrac{1}{20}}=\dfrac{\dfrac{200-10}{200}}{\dfrac1{20}}=\frac{\dfrac{19}{20}}{\dfrac1{20}}=19\end{align}