The given expression is:

$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{n^2 + k}$

Step-by-step solution:

1. Rewriting the sum:
   For each fixed $n$, the sum is given by:

   $S_n = \sum_{k=1}^{n} \frac{1}{n^2 + k}$

   We want to evaluate the limit of this sum as $n \to \infty$.

2. Approximation for large $n$:
   When $n$ is large, for each $k$, we can approximate the denominator $n^2 + k$ by $n^2$ since $k$ is much smaller than $n^2$. Thus,

   $\frac{1}{n^2 + k} \approx \frac{1}{n^2}$

   This approximation allows us to simplify the sum:

   $S_n \approx \sum_{k=1}^{n} \frac{1}{n^2} = \frac{1}{n^2} \sum_{k=1}^{n} 1 = \frac{n}{n^2} = \frac{1}{n}$

   As $n \to \infty$, $\frac{1}{n} \to 0$. This suggests that the limit of the sum might approach 0.

3. Improving the approximation:
   To get a more precise result, observe that the sum resembles a Riemann sum. We rewrite the sum in terms of an integral approximation. Notice that:

   $S_n = \sum_{k=1}^{n} \frac{1}{n^2} \cdot \frac{1}{1 + \frac{k}{n^2}} \approx \frac{1}{n} \sum_{k=1}^{n} \frac{1}{1 + \frac{k}{n^2}}$

   As $n \to \infty$, this sum behaves like the Riemann sum for the integral of the function $f(x) = \frac{1}{x^2}$ over an interval.

   Thus, the sum tends to zero, leading to the result:

   $\lim_{n \to \infty} S_n = 0$

Final Answer:

$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{n^2 + k} = 0$

Would you like further clarification or more details?

Here are some related questions to explore:

1. How do Riemann sums help approximate sums like this?
2. What if the denominator had a different dependence on $n$ (e.g., $n^3 + k$)?
3. How would the result change if the summation limit $n$ was replaced with another variable?
4. Can we apply the same method to a more complex summation involving higher powers of $n$?
5. What is the physical or geometric interpretation of such limits?

Tip: Recognizing when a sum can be approximated by an integral is a powerful technique in calculus, especially for large $n$.