Inequality proof by induction, what to do next in the step I have to prove that for $n = 1, 2...$ it holds: $2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}}$
Base: For $n = 1$ holds, because $2\sqrt{2}-2 < 1$
Step: assume holds for $n_0$.
$2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n + 1}}$. But I do not know what to do next? How this can be proved?

\begin{align}
\frac1{\sqrt{n+1}}+2\sqrt{n+1}-2&=\frac{1+2(n+1)}{\sqrt{n+1}}-2\\
&=\frac{\sqrt{4n^2+12n+9}}{\sqrt{n+1}}-2\\
&>\frac{\sqrt{4n^2+12n+8}}{\sqrt{n+1}}-2\\
&=\frac{2\sqrt{n^2+3n+2}}{\sqrt{n+1}}-2\\
&=\frac{2\sqrt{(n+2)(n+1)}}{\sqrt{n+1}}-2\\
&=2\sqrt{n+2}-2
\end{align}
