Question

\[n - \sqrt{n ^{2} - n} \] Covergent or Divergent?

Answer 1

Convergent. Converges to 0:
\[n-\sqrt{n(n-1)}\]
For extremely large values of n, the difference between n and n-1 is negligible, so basically you have:
\[\sqrt{n(n-1)} \approx \sqrt{n(n)} = \sqrt{n^{2}} = n\]
Thus you have n - n = 0