Deduce a sum by 2% and make it a perfect square!

$\left[\large \left(\Large{\sum_{n=1}^{99}} \frac1{\displaystyle \sum_{m=1}^n m} \right) - 0.02\right]^{\Large \frac{1}{2}} = \ ?$ Then we can proceed as, $\left[\large \left(\Large{\sum_{n=1}^{99}} \frac1{n(n+1)/2} \right) - 0.02\right]^{\Large \frac{1}{2}}$ (as sum of first n natural numbers is $n(n+1)/2$ )

Then, $\left[\large \left(\Large{\sum_{n=1}^{99}} \frac2{n(n+1)} \right) - 0.02\right]^{\Large \frac{1}{2}}$ We can write the fraction as, $\frac{A}{(n)}+\frac{B}{(n+1)}$ $\frac{A(n+1)+B(n)}{(n)(n+1)}$ So we can compare as follows, $\frac2{n(n+1)}=\frac{A(n+1)+B(n)}{(n)(n+1)}$ $2=A(n+1)+B(n)$ If $n=0$ then, $A=2$ If $n=-1$ then, $B=-2$ So we can write the fraction as, $\left[\large \left(\Large{\sum_{n=1}^{99}} \left(\frac2{(n)}-\frac2{(n+1)}\right)\right) - 0.02\right]^{\Large \frac{1}{2}}$
$\left[\large \left(\Large{\sum_{n=1}^{99}}\frac2{(n)}-\Large{\sum_{n=1}^{99}}\frac2{(n+1)}\right) - 0.02\right]^{\Large \frac{1}{2}}$ Now, $\left[\large \left(\frac{2}{1}+\frac{2}{2}+\frac{2}{3}+\cdots+\frac{2}{99}\right)-\left(\frac{2}{2}+\frac{2}{3}+\frac{2}{4}+\cdots+\frac{2}{100}\right)- 0.02\right]^{\Large \frac{1}{2}}$ So all the terms from $\frac{2}{2}$ to $\frac{2}{99}$ will cancel out, The remaining part will be, $\left[\large \left(\frac{2}{1}-\frac{2}{100}\right)- 0.02\right]^{\Large \frac{1}{2}}$ $\left(\large 2-0.02- 0.02\right)^{\Large \frac{1}{2}}$ $\left(\large 2- 0.04\right)^{\Large \frac{1}{2}}$ $\left( 1.96\right)^{\Large \frac{1}{2}}$ $1.4$ So, the answer is $1.4$ .