The problem asks to show that the sequence defined by

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

converges to $\frac{1}{\sqrt{2}}$ as $n \to \infty$.

Step-by-step approach:

1. Rewrite each term of the sum: The $k$-th term of the sum is $\frac{1}{\sqrt{2n^2 + k}}$. For large $n$, we can approximate $\sqrt{2n^2 + k}$ using a first-order approximation (for large $n$):

   $\sqrt{2n^2 + k} \approx n\sqrt{2}\left(1 + \frac{k}{2n^2}\right)^{1/2}.$ Using the binomial expansion for small $\frac{k}{2n^2}$, we get: $\sqrt{2n^2 + k} \approx n\sqrt{2} + \frac{k}{\sqrt{2}n}.$

2. Approximate the term $\frac{1}{\sqrt{2n^2 + k}}$: $\frac{1}{\sqrt{2n^2 + k}} \approx \frac{1}{n\sqrt{2}}\left(1 - \frac{k}{2n^2}\right).$ So, each term is approximately: $\frac{1}{\sqrt{2n^2 + k}} \approx \frac{1}{n\sqrt{2}} - \frac{k}{2n^3\sqrt{2}}.$

3. Sum the terms: The sequence $a_n$ is the sum of these terms: $a_n \approx \sum_{k=1}^{n} \left(\frac{1}{n\sqrt{2}} - \frac{k}{2n^3\sqrt{2}}\right).$ Breaking this into two sums: $a_n \approx \frac{1}{n\sqrt{2}} \sum_{k=1}^{n} 1 - \frac{1}{2n^3\sqrt{2}} \sum_{k=1}^{n} k.$ The first sum is simply $n$, and the second sum is the sum of the first $n$ integers, which is $\frac{n(n+1)}{2}$. So: $a_n \approx \frac{1}{\sqrt{2}} - \frac{n(n+1)}{4n^3\sqrt{2}}.$

4. Simplify the second term: The second term simplifies to: $\frac{n(n+1)}{4n^3\sqrt{2}} = \frac{1}{4n\sqrt{2}} + \frac{1}{4n^2\sqrt{2}}.$ As $n \to \infty$, both of these terms go to zero, so: $a_n \to \frac{1}{\sqrt{2}}.$

Thus, the sequence $a_n$ converges to $\frac{1}{\sqrt{2}}$ as $n \to \infty$.

Let me know if you'd like more details or clarification!

Here are some related questions to extend your understanding:

1. How does the binomial expansion help simplify the sequence?
2. What happens if we take more terms in the approximation of $\sqrt{2n^2 + k}$?
3. Could you approximate this sequence using integrals instead of summation?
4. How does the convergence speed of the sequence depend on $n$?
5. What methods are used to prove the convergence of more complex sequences?

Tip: Always check if you can approximate sums with integrals when dealing with large $n$.