Using the definite integral, find the limit of the sum

Using the definite integral, find the limit of the sum: $\lim_{n \rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\cdots+\frac{2n-1}{n^2}\right)$.

We will express this sum as a Riemann sum and find its limit using the definite integral.

Step 1: Rewrite the sum

The sum can be written as:

\[S_n = \sum_{k=1}^{2n-1} \frac{k}{n^2} = \frac{1}{n^2} \sum_{k=1}^{2n-1} k\]

Alternatively, we can write:

\[S_n = \sum_{k=1}^{2n-1} \frac{k}{n^2} = \sum_{k=1}^{2n-1} \frac{k}{n} \cdot \frac{1}{n}\]

Step 2: Recognize as a Riemann sum

Let $\Delta x = \frac{1}{n}$ and $x_k = \frac{k}{n}$ where $k = 1, 2, \ldots, 2n-1$.

Then:

\[S_n = \sum_{k=1}^{2n-1} \frac{k}{n} \cdot \frac{1}{n} = \sum_{k=1}^{2n-1} x_k \cdot \Delta x\]

This is a Riemann sum for the function $f(x) = x$ over an interval.

Step 3: Determine the interval

The first term corresponds to $x_1 = \frac{1}{n}$ and the last term corresponds to $x_{2n-1} = \frac{2n-1}{n}$.

As $n \to \infty$: $x_1 = \frac{1}{n} \to 0$ and $x_{2n-1} = \frac{2n-1}{n} = 2 - \frac{1}{n} \to 2$.

Therefore, this is a Riemann sum for $f(x) = x$ over the interval $[0, 2]$.

Step 4: Take the limit

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

\[\lim_{n \to \infty} S_n = \int_0^2 x \, dx\]

Step 5: Evaluate the integral

The antiderivative of $x$ is $\frac{x^2}{2}$:

\[\int_0^2 x \, dx = \left[\frac{x^2}{2}\right]_0^2\]

\[= \frac{2^2}{2} - \frac{0^2}{2}\]

\[= \frac{4}{2} - 0\]

\[= 2\]

Alternative verification using the sum formula:

We can verify this result directly using the formula for the sum of the first $m$ positive integers:

\[\sum_{k=1}^{m} k = \frac{m(m+1)}{2}\]

For $m = 2n-1$:

\[S_n = \frac{1}{n^2} \sum_{k=1}^{2n-1} k = \frac{1}{n^2} \cdot \frac{(2n-1)(2n)}{2}\]

\[= \frac{(2n-1)(2n)}{2n^2} = \frac{2n(2n-1)}{2n^2}\]

\[= \frac{4n^2 - 2n}{2n^2}\]

\[= \frac{2n^2 - n}{n^2} = 2 - \frac{1}{n}\]

Taking the limit:

\[\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(2 - \frac{1}{n}\right) = 2 - 0 = 2\]

This confirms our integral result!

Geometric interpretation:

The sum $\frac{1}{n^2}+\frac{2}{n^2}+\cdots+\frac{2n-1}{n^2}$ represents a Riemann sum approximation of the area under the line $y = x$ from $x = 0$ to $x = 2$.

This area forms a right triangle with base 2 (from $x=0$ to $x=2$) and height 2 (at $x=2$), so:

\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2\]

As $n$ increases, the rectangles in the Riemann sum become thinner and more numerous, converging to the exact area of the triangle.

Final answer:

\[\boxed{\lim_{n \rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\cdots+\frac{2n-1}{n^2}\right) = 2}\]