1. **State the problem:** We need to test the convergence of the series with general term $a_n = \sqrt{n^2 + 1} - n$. 2. **Recall the convergence test:** A necessary condition for a series $\sum a_n$ to converge is that $a_n \to 0$ as $n \to \infty$. 3. **Analyze the general term:** Simplify $a_n$ by rationalizing: $$ a_n = \sqrt{n^2 + 1} - n = \frac{(\sqrt{n^2 + 1} - n)(\sqrt{n^2 + 1} + n)}{\sqrt{n^2 + 1} + n} = \frac{n^2 + 1 - n^2}{\sqrt{n^2 + 1} + n} = \frac{1}{\sqrt{n^2 + 1} + n}. $$ 4. **Evaluate the limit:** As $n \to \infty$, $\sqrt{n^2 + 1} \approx n$, so $$ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{\sqrt{n^2 + 1} + n} \approx \lim_{n \to \infty} \frac{1}{n + n} = \lim_{n \to \infty} \frac{1}{2n} = 0. $$ 5. **Check for convergence:** Since $a_n \to 0$, the necessary condition is satisfied, but we must check if the series converges. 6. **Compare with a known series:** Note that $a_n \approx \frac{1}{2n}$ for large $n$, and the series $\sum \frac{1}{n}$ diverges (harmonic series). 7. **Conclusion:** By the comparison test, since $a_n$ behaves like $\frac{1}{2n}$ and $\sum \frac{1}{n}$ diverges, the series $\sum a_n$ diverges. **Final answer:** The series with general term $\sqrt{n^2 + 1} - n$ diverges.