$\frac{A}{B}$ for $A=\frac1{1\cdot2}+\frac1{3\cdot4}+\dots+\frac1{21\cdot22}$ and $B=\frac1{12\cdot22}+\frac1{13\cdot21}+\dots+\frac1{22\cdot12}$

Question

Problem :

Let :

$A=\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{21\cdot22}$

And

$B=\dfrac{1}{12\cdot22}+\dfrac{1}{13\cdot21}+...+\dfrac{1}{22\cdot12}$

Then find the value of : $\dfrac{A}{B}$

My try :

$A=\sum_{n=1}^{11}\dfrac{1}{(2n-1)(2n)}$

Wolfram alpha give me $A=\dfrac{156188887}{2327925960}$

But about B I'm not sure .

$B=\sum_{n=0}^{10}\dfrac{1}{(12+n)(22-n)}=\dfrac{47497}{554400}$

But I don't know how I calculated the summation in fast way ? And I'm searching a simple method

Answer 1

I don't know if it helps but python gives $\frac{A}{B} = 17$

>>> from fractions import Fraction as f
>>> A = sum(f(1, i * (i + 1)) for i in range(1, 22, 2))
>>> B = sum(f(1, i * (34 - i)) for i in range(12, 23))
>>> A / B
Fraction(17, 1)

Answer 2

Hint:

Express $A$ and $B$ in terms of the harmonic numbers: $$H_n = \frac11 + \frac12 + \frac13 + \cdots + \frac1n \quad (n \geqslant 1).$$

It might also help - but this is such a weak hint, there is no need to hide it! - to consider the more general problem of computing a simple form for $\frac{A}{B},$ where for some positive integer $m:$ \begin{gather*} A = \sum_{r=1}^m\frac1{(2r - 1)(2r)}, \\ B = \sum_{r=1}^m\frac1{(m + r)(2m - r + 1)}. \end{gather*}

Answer 3

Using partial fractions decompositions \begin{align} \frac{1}{2n(2n-1)}&=\frac{1}{2n-1}-\frac{1}{2n}\\ \frac{1}{(11+n)(23-n)}&=\frac{1}{34}\left(\frac{1}{11+n}+\frac{1}{23-n}\right) \end{align} we have \begin{align} A&=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{2}\right)+\dots+\left(\frac{1}{21}-\frac{1}{22}\right)\\ &=\left(\frac{1}{1}+\frac{1}{3}+\dots+\frac{1}{21}\right)-\left(\frac{1}{2}+\frac{1}{4}+\dots+\frac{1}{22}\right)\\ &=\left(\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{22}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\dots+\frac{1}{22}\right)\\ &=\left(\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{22}\right)-\left(\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{11}\right)\\ &=\left(\frac{1}{12}+\frac{1}{13}+\dots+\frac{1}{22}\right) \end{align} and \begin{align} 34\cdot B&=\left(\frac{1}{12}+\frac{1}{22}\right)+\left(\frac{1}{13}+\frac{1}{21}\right)+\dots+\left(\frac{1}{22}+\frac{1}{12}\right)\\ &=2\cdot\left(\frac{1}{12}+\frac{1}{13}+\dots+\frac{1}{22}\right)\\ &=2\cdot A \end{align} so $A/B=17$.