Proving that $ \ln\left(\frac{49}{50}\right)<\sum^{98}_{k=1}\int^{k+1}_{k}\frac{k+1}{x(x+1)}dx<\ln(99)$

If $\displaystyle I = \sum^{98}_{k=1}\int^{k+1}_{k}\frac{k+1}{x(x+1)}dx.$ Then prove that $\displaystyle \ln\left(\frac{49}{50}\right)<I <\ln(99)$

Attempt: $$I = \sum^{98}_{k=1}\int^{k+1}_{k}(k+1)\bigg[\frac{1}{x(x+1)}\bigg]dx = \sum^{98}_{k=1}\int^{k+1}_{k}(k+1)\bigg[\frac{1}{x}-\frac{1}{x+1}\bigg]dx$$

So $$I = \sum^{98}_{k=1}(k+1)\bigg[\ln(x)-\ln(x+1)\bigg]\bigg|_{k}^{k+1}$$

$$I= \sum^{98}_{k=1}(k+1)\bigg[\bigg(\ln(k+1)-\ln(k+2)\bigg)-\bigg(\ln(k)-\ln(k+1)\bigg)\bigg]$$

$$ \sum^{98}_{k=1}(k+1)\ln(k+1)-k\ln(k)-\sum^{98}_{k=1}(k+1)\ln(k+2)-k\ln(k+1)+\sum^{98}_{k=1}\ln(k+1)-\ln(k)$$

So we have $$I = \ln(2)+\ln \bigg(\frac{99}{100}\bigg)^{100}$$

could some help me how to prove $$\ln\left(\frac{49}{50}\right)<I <\ln(99)$$

$$\begin{eqnarray*} I &=& \sum_{k=1}^{98}(k+1)\left[2\log(k+1)-\log(k)-\log(k+2)\right]\\&\stackrel{\text{SBP}}{=}& \log(2) + 100\log(99)-99\log(100)\end{eqnarray*}$$ hence by exponentiation it is enough to show that $$ 1-\frac{1}{50} < 2 \frac{99^{100}}{100^{99}} < 99 $$ and by Bernoulli's inequality $$ e^I = 200\left(1-\frac{1}{100}\right)^{100} \approx \frac{200}{e} $$ so the proof is straightforward.

Note that, for $x\in[k,k+1]$, $$ k<x<k+1\Rightarrow\frac{1}{k+1}<\frac{1}{x}<\frac{1}{k}, \frac{1}{k+2}<\frac{1}{x+1}<\frac{1}{k+1}$$ and hence \begin{eqnarray} I &=& \sum^{98}_{k=1}\int^{k+1}_{k}\frac{k+1}{x(x+1)}dx\\ &\le&\sum^{98}_{k=1}\int^{k+1}_{k}\frac{1}{x}dx\\ &=&\sum^{98}_{k=1}[\ln(k+1)-\ln(k)]\\ &=&\ln99 \end{eqnarray} and \begin{eqnarray} I&=&\sum^{98}_{k=1}\int^{k+1}_{k}\frac{k+1}{x(x+1)}dx\\ &\ge&\sum^{98}_{k=1}\int^{k+1}_{k}\frac{1}{x+1}dx\\ &=&\sum^{98}_{k=1}[\ln(k+2)-\ln(k+1)]\\ &=&\ln55>\ln(\frac{49}{50}). \end{eqnarray}