$$\lim_{n \to \infty} \frac{1}{1+n} + \frac{1}{2+n} + \frac{1}{3+n} + \ldots + \frac{1}{2n}$$ is equal to:

We need to evaluate $$\lim_{n \to \infty} \left(\frac{1}{1+n} + \frac{1}{2+n} + \frac{1}{3+n} + \ldots + \frac{1}{2n}\right)$$.

Write the sum in sigma notation.

$$ S_n = \sum_{k=1}^{n} \frac{1}{k+n} $$

Convert to a Riemann sum.

Factor out $$\frac{1}{n}$$ from each term:

$$ S_n = \sum_{k=1}^{n} \frac{1}{n} \cdot \frac{1}{\frac{k}{n} + 1} $$

This is a Riemann sum for the integral $$\int_0^1 \frac{1}{x+1}\, dx$$, where the interval $$[0, 1]$$ is divided into $$n$$ equal subintervals of width $$\Delta x = 1/n$$, and the sample points are $$x_k = k/n$$.

As $$n \to \infty$$, the Riemann sum converges to the definite integral:

$$ \lim_{n \to \infty} S_n = \int_0^1 \frac{dx}{x+1} $$

Evaluate the integral.

Using the standard integral $$\int \frac{dx}{x+1} = \ln|x+1| + C$$:

$$ \int_0^1 \frac{dx}{x+1} = \left[\ln(x+1)\right]_0^1 = \ln(1+1) - \ln(0+1) = \ln 2 - \ln 1 = \ln 2 $$

Therefore, the limit is $$\log_e 2$$.

The correct answer is Option 2: $$\log_e 2$$.