The image presents a limit problem involving a sum of fractions:

$\lim_{n \to \infty} \left( \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \right)$

To solve this, we can recognize that the given sum is a partial sum of the harmonic series shifted by $n$. To find the limit, we can use the properties of integrals to approximate the sum. Here's the step-by-step solution:

1. Rewrite the Sum: The given sum can be written in summation notation: $\sum_{k=1}^{n} \frac{1}{n+k}$

2. Approximate the Sum with an Integral: As $n$ becomes large, the sum can be approximated by an integral. Specifically, we approximate the sum of $f(x)$ at discrete points by the integral of $f(x)$: $\sum_{k=1}^{n} \frac{1}{n+k} \approx \int_{n+1}^{2n} \frac{1}{x} \, dx$

3. Evaluate the Integral: The integral of $\frac{1}{x}$ is $\ln|x|$. So, we integrate from $n+1$ to $2n$: $\int_{n+1}^{2n} \frac{1}{x} \, dx = \ln(2n) - \ln(n+1)$

4. Simplify the Expression: Using properties of logarithms: $\ln(2n) - \ln(n+1) = \ln\left(\frac{2n}{n+1}\right)$

   As $n \to \infty$, the fraction $\frac{2n}{n+1}$ approaches 2: $\lim_{n \to \infty} \ln\left(\frac{2n}{n+1}\right) = \ln(2)$

Thus, the limit is: $\lim_{n \to \infty} \left( \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \right) = \ln(2)$

Would you like more details on the steps or have any other questions?

Here are some additional questions you might consider:

1. Can you explain why the sum can be approximated by an integral?
2. What other methods can be used to approximate sums of this type?
3. How does this approximation change if the range of the sum is different?
4. What are some applications of these types of limits?
5. Can you provide an example of a similar problem with a different function?

Tip: Approximating sums with integrals is a powerful technique, especially when dealing with large $n$. It helps to convert discrete problems into continuous ones for easier evaluation.