How do I show that $\sum_{i = 1}^n \frac 1{\sqrt{a_n}} \lt \frac {\sqrt 3}6$ for $a_n = 4n(4n + 1)(4n + 2)$? Let $a_n = 4n(4n + 1)(4n + 2)$, show that $$\sum_{i = 1}^n \frac 1{\sqrt{a_i}} \lt \frac {\sqrt 3}6 \quad \forall n \in \mathbb{N}^+.$$
I know I need to find an upper bound for $1/\sqrt{a_n}$ but I can't see how, especially with the square root. Any hints will be appreciated!

$$
\begin{align}
\frac12\sum_{k=1}^\infty\left(\frac1{\sqrt{4k-1}}-\frac1{\sqrt{4k+3}}\right)
&=\sum_{k=1}^\infty\frac2{\sqrt{4k-1}\sqrt{4k+3}\left(\sqrt{4k-1}+\sqrt{4k+3}\right)}\tag{1}\\
&\ge\sum_{k=1}^\infty\frac1{\sqrt{4k-1}\sqrt{4k+3}\sqrt{4k+1}}\tag{2}\\
&\ge\sum_{k=1}^\infty\frac1{\sqrt{4k}\sqrt{4k+2}\sqrt{4k+1}}\tag{3}
\end{align}
$$
Explanation:
$(1)$: arithmetic
$(2)$: concavity of $\sqrt{x}$ says that $\frac{\sqrt{x}+\sqrt{y}}2\le\sqrt{\frac{x+y}2}$
$(3)$: $(4k-1)(4k+3)\le4k(4k+2)$ by expanding
This says that
$$
\begin{align}
\sum_{k=1}^\infty\frac1{\sqrt{4k(4k+1)(4k+2)}}
&\le\frac12\sum_{k=1}^\infty\left(\frac1{\sqrt{4k-1}}-\frac1{\sqrt{4k+3}}\right)\\
&=\frac12\frac1{\sqrt3}\\
&=\frac{\sqrt3}6\tag{4}
\end{align}
$$

Motivation
Since the terms of the sum are $\sim\frac18k^{-3/2}$, it is often useful to consider a telescoping series where the terms are differences of something $\sim\frac14k^{-1/2}$ because such a difference is $\sim\frac18k^{-3/2}$.
