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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 \(n \to \infty\).

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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 \(n\).

Show that the sequence a_n = 1/sqrt(2n^2 + 1) + 1/sqrt(2n^2 + 2) + ... + 1/sqrt(2n^2 + n) converges to 1/sqrt(2).

Learn how to prove that the sequence defined by a_n = 1/sqrt(2n^2 + 1) + 1/sqrt(2n^2 + 2) + ... + 1/sqrt(2n^2 + n) converges to 1/sqrt(2). This problem involves key mathematical concepts such as sequences, series, limits, and binomial approximations. Suitable for undergraduate students.

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