1. **Problem statement:** We have the sequence defined by $$u_n = \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}$$ for all $n \geq 1$. We want to show that $$\frac{1}{2} \leq u_n \leq 1$$ for all $n$ in $\mathbb{N}^*$. 2. **Understanding the sum:** The term $u_n$ is the sum of the reciprocals of integers from $n+1$ to $2n$. This can be written as $$u_n = \sum_{k=n+1}^{2n} \frac{1}{k}.$$ 3. **Using integral comparison:** Recall that the function $f(x) = \frac{1}{x}$ is positive and decreasing for $x > 0$. For such functions, the integral test gives inequalities relating sums and integrals: - For decreasing $f$, $$\int_{m}^{M+1} f(x) \, dx \leq \sum_{k=m}^M f(k) \leq \int_{m-1}^M f(x) \, dx.$$ 4. **Apply to our sum:** Set $m = n+1$ and $M = 2n$. Then, $$\int_{n+1}^{2n+1} \frac{1}{x} \, dx \leq u_n \leq \int_n^{2n} \frac{1}{x} \, dx.$$ 5. **Evaluate the integrals:** The integral of $1/x$ is $\ln x$, so $$\int_a^b \frac{1}{x} \, dx = \ln b - \ln a = \ln \left(\frac{b}{a}\right).$$ Thus, $$\ln \left(\frac{2n+1}{n+1}\right) \leq u_n \leq \ln \left(\frac{2n}{n}\right) = \ln 2.$$ 6. **Simplify the lower bound:** Note that $$\frac{2n+1}{n+1} = 2 - \frac{1}{n+1}.$$ Since $n \geq 1$, $\frac{1}{n+1} \leq \frac{1}{2}$, so $$2 - \frac{1}{n+1} \geq 2 - \frac{1}{2} = \frac{3}{2} = 1.5.$$ Therefore, $$\ln \left(\frac{2n+1}{n+1}\right) \geq \ln \left(\frac{3}{2}\right) \approx 0.405.$$ 7. **Compare bounds to $\frac{1}{2}$ and $1$:** We have $$0.405 \lesssim u_n \leq \ln 2 \approx 0.693.$$ Since $0.405 < \frac{1}{2}$, this is not enough to prove the lower bound $\frac{1}{2} \leq u_n$ directly from this integral estimate. 8. **Alternative approach for the lower bound:** Consider the sum $u_n$ as the sum of $n$ terms, each term $\frac{1}{k}$ with $k$ from $n+1$ to $2n$. Since $k \leq 2n$, each term is at least $\frac{1}{2n}$. Thus, $$u_n = \sum_{k=n+1}^{2n} \frac{1}{k} \geq n \times \frac{1}{2n} = \frac{1}{2}.$$ 9. **Summary:** We have shown - Lower bound: $$u_n \geq \frac{1}{2}.$$ - Upper bound: $$u_n \leq \ln 2 < 1.$$ Since $\ln 2 \approx 0.693 < 1$, the upper bound $u_n \leq 1$ holds. **Final conclusion:** For all $n \geq 1$, $$\frac{1}{2} \leq u_n \leq 1.$$