Question

Answer it
And do not spam​

Answer 1

Given Question

Find the value of

$\rm :\longmapsto\:\dfrac{\displaystyle\sum_{r=1}^n \frac{1}{r} }{\displaystyle\sum_{k=1}^n \frac{k}{(2n - 2k + 1)(2n - k + 1)} }$

$\green{\large\underline{\sf{Solution-}}}$

Given expression is

$\rm :\longmapsto\:\dfrac{\displaystyle\sum_{r=1}^n \frac{1}{r} }{\displaystyle\sum_{k=1}^n \frac{k}{(2n - 2k + 1)(2n - k + 1)} }$

Let solve the numerator and denominator separately.

So, Let consider numerator

$\rm :\longmapsto\:\displaystyle\sum_{r=1}^n \frac{1}{r}$

Let assume that

$\rm :\longmapsto\:a = \displaystyle\sum_{r=1}^n \frac{1}{r}$

$\bf \implies\:a = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + - - - + \dfrac{1}{n} - - - (1)$

Now, Consider Denominator

$\rm :\longmapsto\:\displaystyle\sum_{k=1}^n \frac{k}{(2n - 2k + 1)(2n - k + 1)}$

Let us assume that

$\rm :\longmapsto\:b = \displaystyle\sum_{k=1}^n \frac{k}{(2n - 2k + 1)(2n - k + 1)}$

can be rewritten as

$\rm :\longmapsto\:b = \displaystyle\sum_{k=1}^n \frac{2k - k}{(2n - 2k + 1)(2n - k + 1)}$

$\rm :\longmapsto\:b = \displaystyle\sum_{k=1}^n \frac{2k - k + 2n - 2n + 1 - 1}{(2n - 2k + 1)(2n - k + 1)}$

$\rm :\longmapsto\:b = \displaystyle\sum_{k=1}^n \frac{(2n - k + 1) - (2n - k + 1)}{(2n - 2k + 1)(2n - k + 1)}$

$\rm \:  =  \: \displaystyle\sum_{k=1}^n\bigg[\dfrac{1}{2n - 2k + 1} - \dfrac{1}{2n - k + 1} \bigg]$

$\rm =\bigg[\dfrac{1}{2n - 1}+ - -+ \dfrac{1}{3} + \dfrac{1}{1}\bigg] - \bigg[\dfrac{1}{2n} + - - \dfrac{1}{n + 2} + \dfrac{1}{n + 1} \bigg]$

Now, Consider,

$\rm :\longmapsto\:a - b$

$\rm=\bigg[\dfrac{1}{1}+\dfrac{1}{2} + - - + \dfrac{1}{n}+\dfrac{1}{n + 1} --\dfrac{1}{2n} \bigg] - \bigg[1 + \dfrac{1}{3} + \dfrac{1}{5} + - - + \dfrac{1}{2n - 1} \bigg]$

$\rm \:  =  \: \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + - - + \dfrac{1}{2n}$

$\rm \:  =  \:\dfrac{1}{2} \bigg[ \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + - - + \dfrac{1}{n} \bigg]$

$\rm \:  =  \: \dfrac{a}{2}$

So,

$\rm \implies\:a - b =  \: \dfrac{a}{2}$

$\rm \implies\:a \: -   \: \dfrac{a}{2} = b$

$\rm \implies\:\dfrac{a}{2} = b$

$\rm \implies\:\dfrac{a}{b} = 2$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{\displaystyle\sum_{r=1}^n \frac{1}{r} }{\displaystyle\sum_{k=1}^n \frac{k}{(2n - 2k + 1)(2n - k + 1)} } = 2 \: }}$