Why is the series with the general term $\sqrt{n^2 + 1} - n$ divergent?

Question

Test the series whose general term is: $$\sqrt{n^2 + 1} - n$$

This is from Higher Algebra by Henry Sinclair. I'm struggling to understand how this series is divergent. If you do the following expansion: $$\sqrt{n^2 + 1} - n$$ $$= n\biggl(1 + \frac{1}{n^2}\biggr)^\frac{1}{2} - n$$ $$= n + \frac{1}{2n} - \frac{1}{8n^3} + ... - n$$ $$= \frac{1}{2n} - \frac{1}{8n^3} + ... $$

The remaining series appears to converge to zero as $n$ approaches infinity.

Their explanation is similar to mine, but at the end they say that the above series approaches $\frac{1}{2n}$. From there they directly say that the series is divergent. Earlier in the chapter, it's proved that the expansion of $(1 + x)^n$ is convergent for $x < 1$. In this case, $\frac{1}{n^2} < 1$ as $n$ approaches infinity, so I would assume that the series converges.

The solution to the problem is given here (number 17)

Any help is appreciated. Thanks.

Answer 1

Hint: $\sqrt{n^2+1} - n> \dfrac{1}{3n}$ and the latter series is harmonic which is a popular diverging series.

Answer 2

To find the nature of a series, we compare the general term to that of a known series (convergent or divergent)

For example the Riemann series are used to compare:

$\frac{1}{n^{1/2}}$ (divergent), $\frac{1}{n}$ (divergent), $\frac{1}{n^{2}}$ (convergent), …

Here

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

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

$>\frac{1}{4}.\frac{1}{n}$

We can also compare general terms with equivalents. But we also need to know reference series.