Summation of Definite Integrals Bound Estimation

To find the bounds for the given integral summation $I = \sum_{k=1}^{98} \int_{k}^{k+1} \frac{k+1}{x(x+1)} \, dx$ we can analyze the integrand $\frac{k+1}{x(x+1)}$ on the interval $x \in (k, k+1)$.

Step 1: Finding the Upper Bound

For any $x \in (k, k+1)$, we have $x > k$, which implies $x + 1 > k + 1$. Therefore: $\frac{k+1}{x+1} < 1$

Multiplying both sides by $\frac{1}{x}$ (since $x > 0$): $\frac{k+1}{x(x+1)} < \frac{1}{x}$

Integrating both sides from $x = k$ to $x = k+1$: $\int_{k}^{k+1} \frac{k+1}{x(x+1)} \, dx < \int_{k}^{k+1} \frac{1}{x} \, dx = [\ln x]_{k}^{k+1} = \ln(k+1) - \ln(k)$

Summing this inequality from $k = 1$ to $k = 98$: $I < \sum_{k=1}^{98} \left( \ln(k+1) - \ln(k) \right)$

The right-hand side is a telescoping sum: $\sum_{k=1}^{98} \left( \ln(k+1) - \ln(k) \right) = \ln(99) - \ln(1) = \log_e 99$

Thus: $I < \log_e 99$ This proves that Option B is correct (and Option A is incorrect).

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Step 2: Finding the Lower Bound

For any $x \in (k, k+1)$, we also have $x < k+1$, which implies: $\frac{k+1}{x} > 1$

Multiplying both sides by $\frac{1}{x+1}$: $\frac{k+1}{x(x+1)} > \frac{1}{x+1}$

Integrating both sides from $x = k$ to $x = k+1$: $\int_{k}^{k+1} \frac{k+1}{x(x+1)} \, dx > \int_{k}^{k+1} \frac{1}{x+1} \, dx = [\ln(x+1)]_{k}^{k+1} = \ln(k+2) - \ln(k+1)$

Summing this inequality from $k = 1$ to $k = 98$: $I > \sum_{k=1}^{98} \left( \ln(k+2) - \ln(k+1) \right)$

This is again a telescoping sum: $\sum_{k=1}^{98} \left( \ln(k+2) - \ln(k+1) \right) = \ln(100) - \ln(2) = \ln\left(\frac{100}{2}\right) = \log_e 50$

Since $50 > e \approx 2.718$, we have $\log_e 50 > 1 > \frac{49}{50}$.

Thus: $I > \frac{49}{50}$ This proves that Option D is correct (and Option C is incorrect).

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Conclusion

The correct inequalities that hold true are B and D.