Answer :

To find the limit of the given sum as [tex]\( n \)[/tex] approaches infinity, we can transform the discrete sum into an integral, which is often easier to evaluate in such cases.

Given the sum:
[tex]\[ \lim_{n \to \infty} \sum_{r=1}^n \left(\frac{r}{n^2 + r}\right) \][/tex]

1. Rewrite the term inside the summation:

Consider the expression inside the sum:
[tex]\[ \frac{r}{n^2 + r} \][/tex]

For large [tex]\( n \)[/tex], [tex]\( r \)[/tex] will vary from 1 to [tex]\( n \)[/tex]. We divide the numerator and the denominator by [tex]\( n^2 \)[/tex] to get:
[tex]\[ \frac{r}{n^2 + r} = \frac{r/n^2}{1 + r/n^2} = \frac{\frac{r}{n^2}}{1 + \frac{r}{n^2}} \][/tex]

2. Transform the sum into an integral:

We recognize that this sum as [tex]\( n \to \infty \)[/tex] can be approximated by an integral. Define [tex]\( x = \frac{r}{n} \)[/tex], so [tex]\( r = nx \)[/tex]. When [tex]\( r \)[/tex] ranges from 1 to [tex]\( n \)[/tex], [tex]\( x \)[/tex] ranges from [tex]\( \frac{1}{n} \)[/tex] to 1. The sum now becomes:
[tex]\[ \sum_{r=1}^n \left(\frac{\frac{r}{n^2}}{1 + \frac{r}{n^2}}\right) \approx \sum_{r=1}^n \left( \frac{x/n}{1 + x/n} \cdot n \cdot \frac{1}{n} \right) = \sum_{r=1}^n \left( \frac{x/n}{1 + x/n} \cdot \frac{1}{n} \right) \][/tex]

We recognize this as a Riemann sum for the integral:
[tex]\[ \int_{0}^{1} \frac{x}{1 + x} \, dx \][/tex]

3. Evaluate the integral:

Now, we need to evaluate the integral:
[tex]\[ \int_{0}^{1} \frac{x}{1 + x} \, dx \][/tex]

Perform the substitution [tex]\( u = 1 + x \)[/tex], hence [tex]\( du = dx \)[/tex], and when [tex]\( x = 0 \)[/tex], [tex]\( u = 1 \)[/tex]; when [tex]\( x = 1 \)[/tex], [tex]\( u = 2 \)[/tex]:
[tex]\[ \int_{0}^{1} \frac{x}{1 + x} \, dx = \int_{1}^{2} \frac{u-1}{u} \, du = \int_{1}^{2} \left(1 - \frac{1}{u}\right) \, du \][/tex]

Calculate the integral:
[tex]\[ \int_{1}^{2} 1 \, du - \int_{1}^{2} \frac{1}{u} \, du = \left[u \right]_{1}^{2} - \left[\ln|u| \right]_{1}^{2} \][/tex]

Evaluate these terms:
[tex]\[ \left[u \right]_{1}^{2} = 2 - 1 = 1 \][/tex]
[tex]\[ \left[\ln|u| \right]_{1}^{2} = \ln|2| - \ln|1| = \ln 2 - 0 = \ln 2 \][/tex]

Combine these results:
[tex]\[ 1 - \ln 2 \][/tex]

Therefore, the value of the limit is:
[tex]\[ \boxed{1 - \ln 2} \][/tex]