Evaluate `("lim")_(nvecoo)sum_(k=1)^nk/(n^2+k^2)`

Explanation

To evaluate the limit

\[ L = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{n^2 + k^2}, \]

we will follow a step-by-step approach.

### Step 1: Rewrite the expression
We start by rewriting the limit:

\[ L = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{n^2 + k^2}. \]

### Step 2: Factor out \(n^2\) from the denominator
Next, we factor \(n^2\) out of the denominator:

\[ L = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{n^2(1 + \frac{k^2}{n^2})} = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{n^2} \cdot \frac{k}{1 + \frac{k^2}{n^2}}. \]

### Step 3: Change the summation to a Riemann sum
Now, we can express the summation in terms of \(x = \frac{k}{n}\):

\[ L = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{n} \cdot \frac{\frac{k}{n}}{1 + \left(\frac{k}{n}\right)^2} \cdot \frac{1}{n}. \]

This is now in the form of a Riemann sum for the function \(f(x) = \frac{x}{1 + x^2}\) over the interval from \(0\) to \(1\).

### Step 4: Convert the summation to an integral
As \(n\) approaches infinity, the summation converges to the integral:

\[ L = \int_{0}^{1} \frac{x}{1 + x^2} \, dx. \]

### Step 5: Evaluate the integral
To evaluate the integral, we can use the substitution \(u = 1 + x^2\), which gives \(du = 2x \, dx\) or \(dx = \frac{du}{2x}\). The limits change from \(x = 0\) to \(x = 1\) which corresponds to \(u = 1\) to \(u = 2\):

\[ \int \frac{x}{1 + x^2} \, dx = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln |u| + C. \]

Thus, we have:

\[ \int_{0}^{1} \frac{x}{1 + x^2} \, dx = \frac{1}{2} [\ln(2) - \ln(1)] = \frac{1}{2} \ln(2). \]

### Step 6: Final result
Therefore, the limit evaluates to:

\[ L = \frac{1}{2} \ln(2). \]