Dao Trong Luan : \(\dfrac{100}{400}< \dfrac{1}{4}?\)

We have:

\(\dfrac{1}{5};\dfrac{1}{45};\dfrac{1}{117};...\) <=> \(\dfrac{1}{1\cdot5};\dfrac{1}{5\cdot9};\dfrac{1}{9\cdot13};...\)

So the first numbers of denominator of  fraction 100^th of this series is:

\(\left(100-1\right)\cdot4+1=397\)

=> This fraction is \(\dfrac{1}{397\cdot\left(397+4\right)}=\dfrac{1}{397\cdot401}\)

So their sum are:

\(\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+...+\dfrac{1}{397\cdot401}\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{397}-\dfrac{1}{401}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)=\dfrac{1}{4}\cdot\dfrac{400}{401}=\dfrac{100}{401}< \dfrac{100}{400}< \dfrac{1}{4}\)

So the sum of 100 fraction in the series is smaller than \(\dfrac{1}{4}\)

Oh, sorry, I write mistake, sorry

\(\dfrac{100}{400}=\dfrac{1}{4}\)