Square root series convergence or divergence? I'm trying to show whether the series $\sum_{n=1}^{\infty} \sqrt{n+1} - \sqrt{n}$ converges or diverges using the direct comparison test. 
Attempt:
$\sqrt{n+1} - \sqrt{n} = \sqrt{n}\Big(\sqrt{1+ \frac{1}{n}} - 1\Big)$.
Then for positive $t$ we have $ 1 < \sqrt{1+t} < 1 + \frac{t}{2}$, because we have $(1 + \frac{t}{2})^{2} > 1 + t$.
Therefore we have $ 0 < \sqrt{n}\Big(\sqrt{1 + \frac{1}{n}} - 1\Big) < \frac{1}{2 \sqrt{n}}$. 
Now I know $\frac{1}{2 \sqrt{n}}$ diverges. But I'm not sure I can claim the original series diverges since for the Direct comparison test to apply I have to show the original series is greater than a divergent series?
Thanks.
 A: Directly with a known test:
$$\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}\ge\frac1{\sqrt{2n}+\sqrt n}=\frac1{\sqrt2+1}\cdot\frac1{\sqrt n}$$
and the last one is a sequence whose infinite series diverges as a scalar multiple of the divergent series $\;\sum\limits_{n=1}^\infty\cfrac1{\sqrt n}\;$ ...and now use the comparison test for positive series.
Another way: observe that
$$S_k:=\sum_{n=1}^k(\sqrt{n+1}-\sqrt n)=\left(\sqrt2-\sqrt1\right)+\left(\sqrt3-\sqrt2\right)+\ldots+\left(\sqrt{k+1}-\sqrt k\right)=$$
$$=\sqrt{k+1}-1\xrightarrow[k\to\infty]{}\infty$$
and this means the sequence of partial sums of the series diverges to $\;\infty\;$ and thus so does the infinite series.
A: We can use that
$$ \sqrt{n}\left(\sqrt{1+ \frac{1}{n}} - 1\right) \sim \sqrt{n} \cdot\frac1{2n}=\frac1{2\sqrt n}$$
and then conclude by limit comparison test.
As an alternative we can use that
$$\sqrt{1+ \frac{1}{n}} \ge 1+ \frac{1}{2n}-\frac{1}{8n^2}$$
and conclude by direct comparison test.
A: Actually, we have $$\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}\sim_{n\to\infty}\frac1{2\sqrt n}$$
and the latter diverges, so the former does too.
Added:
I'll show why it two series $\sum_n a_n $ and $\sum_n b_n$ have asymptotically equivalent (ultimately) positive terms, and if 
one of them diverges, the other diverges too.
Indeed,suppose $\sum_n b_n$ diverges. By definition of equivalence, $\lim_{n\to\infty}\dfrac{a_n}{b_n}=1$, so for any $\varepsilon>0$, there exists an integer $N_0$ such that
\begin{align}
&&1-\varepsilon<\frac{a_n}{b_n}&<1+\varepsilon&&\text{ for all }\quad n\ge N_0\qquad \\
&\text{whence, as }\; b_n>0, &a_n&>(1-\varepsilon)b_n&&\text{ for all }\quad n\ge N_0,
\end{align}
which by the comparison test, implies that $\sum_n a_n$ diverges, just like the series $(1-\varepsilon)\sum_n b_n$.
