$\displaystyle\sum_{k=1}^{98} \int\limits_{k}^{k+1} \frac{1}{x+1} d x<\displaystyle\sum_{k=1}^{98} \int\limits_{k}^{k+1} \frac{k+1}{x(x+1)} d x<\displaystyle\sum_{k=1}^{98} \int\limits_{k}^{k+1} \frac{d x}{x}$
$\displaystyle\sum_{k=1}^{98}(\ell n(k+2)-\ln (k+1)) < I < \displaystyle\sum_{k=1}^{98}(\ell n(k+1)-\ell n k)$
$\ell n 50 < I < \ell n 99$
$\Rightarrow \frac{49}{50} < I < \ell n 99$