Quadruplets Under One Roof

Note that for all $k \geq 1$ $(k+1)(k+2) - k(k+3) = 2 > 0 \implies \left((k+1)(k+2)\right)^2 > k(k+1)(k+2)(k+3) \implies \frac{1}{(k+1)(k+2)} < \frac{1}{\sqrt{k(k+1)(k+2)(k+3)}}$ $(k+2)(k+3) - k(k+1) > 5k+6 > 0 \implies k(k+1)(k+2)(k+3) > \left(k(k+1)\right)^2 \implies \frac{1}{\sqrt{k(k+1)(k+2)(k+3)}} < \frac{1}{k(k+1)}$ In particular, this shows $\sum_{k=1}^\infty \frac{1}{k(k+1)} - \frac{1}{2} = \sum_{k=1}^\infty \frac{1}{(k+1)(k+2)} < \sum_{k=1}^\infty \frac{1}{\sqrt{k(k+1)(k+2)(k+3)}} < \sum_{k=1}^\infty \frac{1}{k(k+1)}$

Since $\sum_{k=1}^\infty \frac{1}{k(k+1)} = \lim_{N\to\infty} \sum_{k=1}^N \left(\frac{1}{k} - \frac{1}{k+1}\right) = \lim_{N\to\infty} \left(1 - \frac{1}{N+1}\right) = 1,$ the above inequality becomes $\frac{1}{2} = 1 - \frac{1}{2} < \sum_{k=1}^\infty \frac{1}{\sqrt{k(k+1)(k+2)(k+3)}} < 1$ $1 < 2\sum_{k=1}^\infty \frac{1}{\sqrt{k(k+1)(k+2)(k+3)}} < 2$ and therefore $\left\lfloor 2\sum_{k=1}^\infty \frac{1}{\sqrt{k(k+1)(k+2)(k+3)}} \right\rfloor = \boxed{1}$

Note that $k(k+2) = k^2+2k < k^2 +2k + 1 = (k+1)^2$ and $(k+1)(k+3) = k^2 + 4k+3 < k^2+4k+4 = (k+2)^2$ . Therefore,

$\begin{aligned} \frac 1{\sqrt{(k+1)(k+1)(k+2)(k+2)}} < & \frac 1{\sqrt{k(k+1)(k+2)(k+3)}} < \frac 1{\sqrt{k(k)(k+1)(k+1)}} \\ \implies \sum_{k=1}^\infty \frac 1{(k+1)(k+2)} < & \sum_{k=1}^\infty \frac 1{\sqrt{k(k+1)(k+2)(k+3)}} < \sum_{k=1}^\infty \frac 1{k(k+1)} \\ \sum_{k=1}^\infty \left(\frac 1{k+1} - \frac 1{k+2}\right) < & \sum_{k=1}^\infty \frac 1{\sqrt{k(k+1)(k+2)(k+3)}} < \sum_{k=1}^\infty \left(\frac 1k - \frac 1{k+1}\right) \\ \frac 12 < & \sum_{k=1}^\infty \frac 1{\sqrt{k(k+1)(k+2)(k+3)}} < 1 \end{aligned}$

Therefore, $\left \lfloor 2\displaystyle \sum_{k=1}^\infty \frac 1{\sqrt{k(k+1)(k+2)(k+3)}} \right \rfloor = \boxed{1}$ .

Read this first.

$\cfrac{1}{2}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)}}<1$

$1<2\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)}}<2$

$\therefore \lfloor2\displaystyle{\lim_{n\to \infty}} \sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)}}\rfloor=1$

Simply, $\lfloor 2\displaystyle \sum_{k=1}^{\infty} \cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)}}\rfloor=1$