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]
