$$\sum_{n=1}^\infty\frac{\sqrt{\vphantom{A^b}n+1}-\sqrt{n}}{n^x}$$
I multiplied by the conjugate and was able to simplify this to
$$\sum_{n=1}^\infty\frac{1}{n^x(\sqrt{\vphantom{A^b}n+1}+\sqrt{n})}$$
so the series will converge, as the values go to zero as $n$ goes to $\infty$. However, I can't figure out for which $x$ it converges. I want to say it's for all $x>0$, but I'm not sure how to justify this.
Any help appreciated!
Hint: Your rewriting is indeed very useful. By limit comparison or just plain comparison, your series converges iff $\sum \frac{1}{n^{x+1/2}}$ converges. Now use known results about $p$-series. The conclusion should be that we have convergence if and only if $x\gt 1/2$.
If $\;x>\frac12\iff x=\frac12+\epsilon\,,\,\,\epsilon>0\;$ :
$$\frac1{n^x\left(\sqrt{n+1}+\sqrt n\right)}=\frac1{n^{1/2+\epsilon}\left(\sqrt{n+1}+\sqrt n\right)}\le\frac1{n^{1/2+\epsilon}\sqrt n}=\frac1{n^{1+\epsilon}}$$
and we have convergence by comparison.
If $\;x<\frac12\;$
$$\frac1{n^x\left(\sqrt{n+1}+\sqrt n\right)}\ge\frac1{n^{1/2}\left(\sqrt{n+1}+\sqrt n\right)}\ge\frac1{2n^{1/2}(n+1)^{1/2}}\ge\frac12\frac1{\sqrt{n+1}}$$
and the series diverges, again by comparison.
