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!
 A: We may notice that:
$$ \forall n\geq 1,\qquad  \frac{1}{\sqrt{4n(4n+1)(4n+2)}}\leq \frac{1}{2}\left(\frac{1}{\sqrt{4n-1}}-\frac{1}{\sqrt{4n+3}}\right) $$
hence it follows that:
$$ \sum_{n=1}^{N}\frac{1}{\sqrt{4n(4n+1)(4n+2)}} < \left.\frac{1}{2\sqrt{4n-1}}\right|_{n=1}=\color{red}{\frac{1}{2\sqrt{3}}}$$
as wanted. Creative telescoping wins again.
A: $$
\begin{align}
\frac12\sum_{k=1}^\infty\left(\frac1{\sqrt{4k-1}}-\frac1{\sqrt{4k+3}}\right)
&=\sum_{k=1}^\infty\frac2{\sqrt{4k-1}\sqrt{4k+3}\left(\sqrt{4k-1}+\sqrt{4k+3}\right)}\tag{1}\\
&\ge\sum_{k=1}^\infty\frac1{\sqrt{4k-1}\sqrt{4k+3}\sqrt{4k+1}}\tag{2}\\
&\ge\sum_{k=1}^\infty\frac1{\sqrt{4k}\sqrt{4k+2}\sqrt{4k+1}}\tag{3}
\end{align}
$$
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 $\sim\frac18k^{-3/2}$.
A: An attempt :
$$\sum_{k=2}^n \frac{1}{(4k(4k+1)(4k+2))^{1/2}}\le\sum_{k=2}^n \frac{1}{(4k)^{3/2}}\le\sum_{k=2}^\infty \frac{1}{(4k)^{3/2}}\le\int_1^\infty \frac{1}{(4x)^{3/2}}=\frac{1}{4}. $$
We only get that the sum is smaller than $\frac{1}{4}+\frac{1}{\sqrt {120}}$.
A: you can write inequality as $2\sqrt{3}<\sqrt{4n(4n+1)(4n+2)}$ squaring we get it as $12<4n(4n+1)(4n+2)$ this the function is continuously increasing as $n$ is positive. Also $n\in N+$ this base case is $1$ so plugging in we get $120$ this $12<120$ so it's true for akin as function is monotonic and increasing
