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

Given,

$$\lim_{n\to\infty}\left(1+\frac{1+\frac12+\cdots+\frac1n}{n^2}\right)^n$$

Let

$$H_n=1+\frac12+\cdots+\frac1n$$

Then expression becomes

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

Now,

$$H_n\sim\log n$$

Hence,

$$\frac{H_n}{n^2}\to0$$

Using standard result,

$$\left(1+\frac{a_n}{n}\right)^n\to e^L \quad \text{if } a_n\to L$$

Here,

$$a_n=\frac{H_n}{n}\to0$$

Therefore,

$$\left(1+\frac{H_n}{n^2}\right)^n\to e^0$$

$$=1$$

Hence, the required limit is

$$\boxed{1}$$