How does this series diverge?

Question

The series:

$$\sum_{n=0}^{\infty} \sqrt{n^2 +1} -n$$

diverges. Can someone please tell me how this is proven and done.

Answer 1

Hint: $\sqrt{n^2+1} - n = \frac{(\sqrt{n^2+1} - n)(\sqrt{n^2 + 1} + n)}{\sqrt{n^2+1} + n}$.

Answer 2

Notice

$$ \sqrt{n^2+1} - n = \frac{(\sqrt{n^2+1} - n)(\sqrt{n^2+1} + n)}{\sqrt{n^2+1} + n} = \frac{1}{\sqrt{n^2+1} + n}$$

Answer 3

By multiplying top and bottom by $\sqrt{n^2+1}+\sqrt{n}$, we find that the $n$-th term is equal to $$\frac{1}{\sqrt{n^2+1}+n}.$$ This is $\gt \dfrac{1}{\sqrt{n^2+3n^2}+n}$, which is $\frac{1}{3}\cdot\frac{1}{n}$. But we know that $\sum_1^\infty\frac{1}{n}$ diverges.