TITLE: Summation of a series of integrations
QUESTION [6 upvotes]: If $$I = \sum_{k=1}^{98} \int_k^{k+1} \frac{k+1}{x(x+1)}dx,$$ then which option(s) are correct?
$$(a) \,I > \ln 99
\\(b) \,I < \ln 99
\\(c) \,I < \frac{49}{50}
\\(d) \,I > \frac{49}{50}
$$
The correct options are (b) and (d).
My attempt: I tried to simplify the integration and got $(k+1) \ln\bigg|\frac{(k+1)^2}{k(k+2)}\bigg|$
Then the expansion comes as $2 \ln\bigg|\frac{4}{3}\bigg| + 3 \ln\bigg|\frac{9}{8}\bigg| + ...+ 99 \ln\bigg|\frac{99^2}{9800}\bigg|$
Please help me to reach the solution from this step.

REPLY [6 votes]: Note that since $k\le x\le k+1$, then $\frac1{x+1}\le\frac{k+1}{x(x+1)}\le\frac1x$.  Hence, we see that 
$$\log\left(\frac{k+2}{k+1}\right)<\int_k^{k+1}\frac{k+1}{x(x+1)}\,dx< \log\left(\frac{k+1}k\right)$$
Next, noting $\log\left(\frac{k+2}{k+1}\right)=\log(k+2)-\log(k+1)$ and $\log\left(\frac{k+1}k\right)=\log(k+1)-\log(k)$, we find upon summing the telescoping series that 
$$\log(50)<\sum_{k=1}^{98}\int_k^{k+1}\frac{k+1}{x(x+1)}\,dx<\log(99)$$
Finally, given that $\log(x)\ge \frac{x-1}{x}$ we arrive at 
$$\frac{49}{50}<\sum_{k=1}^{98}\int_k^{k+1}\frac{k+1}{x(x+1)}\,dx<\log(99)$$
as expected!