1. **Problem Statement:** Determine whether the series $$\sum_{n=1}^{\infty} \left(\sqrt{n^{2}+1} - n\right)$$ converges. 2. **Recall the Test for Convergence:** For a series $$\sum a_n$$ to converge, the terms $$a_n$$ must approach zero as $$n \to \infty$$. However, this is necessary but not sufficient. We often use comparison or limit comparison tests. 3. **Analyze the general term:** Consider the term $$a_n = \sqrt{n^{2}+1} - n$$. 4. **Simplify the term:** Multiply numerator and denominator by the conjugate to simplify: $$ a_n = \sqrt{n^{2}+1} - n = \frac{(\sqrt{n^{2}+1} - n)(\sqrt{n^{2}+1} + n)}{\sqrt{n^{2}+1} + n} = \frac{n^{2}+1 - n^{2}}{\sqrt{n^{2}+1} + n} = \frac{1}{\sqrt{n^{2}+1} + n}$$ 5. **Estimate the denominator for large $$n$$:** $$\sqrt{n^{2}+1} + n \approx n + n = 2n$$ as $$n \to \infty$$. 6. **Approximate the term for large $$n$$:** $$a_n \approx \frac{1}{2n}$$. 7. **Compare with the harmonic series:** Since $$a_n$$ behaves like $$\frac{1}{2n}$$ for large $$n$$, and $$\sum \frac{1}{n}$$ diverges (harmonic series), by the Limit Comparison Test, our series also diverges. 8. **Conclusion:** The series $$\sum_{n=1}^{\infty} \left(\sqrt{n^{2}+1} - n\right)$$ diverges.