$$\lim_{n \to \infty} \left[\frac{1}{n} + \frac{n}{(n+1)^2} + \frac{n}{(n+2)^2} + \ldots + \frac{n}{(2n-1)^2}\right]$$ is equal to

The given expression is $$\lim_{n \to \infty} \left[\frac{1}{n} + \frac{n}{(n+1)^2} + \frac{n}{(n+2)^2} + \ldots + \frac{n}{(2n-1)^2}\right]$$.

The first term $$\frac{1}{n} = \frac{n}{n^2} = \frac{n}{(n+0)^2}$$, so the entire sum can be written as $$\sum_{k=0}^{n-1} \frac{n}{(n+k)^2}$$.

Factoring out $$\frac{1}{n}$$ from each term: $$\sum_{k=0}^{n-1} \frac{n}{(n+k)^2} = \sum_{k=0}^{n-1} \frac{1}{n} \cdot \frac{n^2}{(n+k)^2} = \sum_{k=0}^{n-1} \frac{1}{n} \cdot \frac{1}{\left(1 + \frac{k}{n}\right)^2}$$.

This is a Riemann sum of the form $$\sum_{k=0}^{n-1} \frac{1}{n} \cdot f\left(\frac{k}{n}\right)$$ where $$f(x) = \frac{1}{(1+x)^2}$$, corresponding to the integral $$\int_0^1 \frac{1}{(1+x)^2}\,dx$$ over the partition $$x_k = \frac{k}{n}$$ for $$k = 0, 1, \ldots, n-1$$.

Evaluating the integral: $$\int_0^1 \frac{1}{(1+x)^2}\,dx = \left[-\frac{1}{1+x}\right]_0^1 = \left(-\frac{1}{2}\right) - \left(-1\right) = -\frac{1}{2} + 1 = \frac{1}{2}$$.

Therefore, the limit equals $$\frac{1}{2}$$.