Ratio of Diverging Series'!

$\mathcal P=\displaystyle\sum_{n=1}^{\infty}\left[\dfrac{\displaystyle\sum_{r=1}^{n}( r+2)}{\displaystyle\sum_{r=1}^n r(r+2)(r+4)}\right]$

Numerator is a sum of an arithmetic progression which is: $\small{\dfrac n2(2\cdot 3+(n-1)1)=\dfrac{n(n+5)}2}$ .

While denominator $(\color{#3D99F6}{**})$ can be written as:

$\sum_{r=1}^{n}(r^3+6r^2+8r)$

Use $\small{\color{#20A900}{\sum r=\dfrac{n(n+1)}2,\sum r^2=\dfrac{n(n+1)(2n+1)}6,\sum r^3=\dfrac{n^2(n+1)^2}4}}$

so that denominator simplifies to :

$\dfrac{n^2(n+1)^2}{4}+n(n+1)(2n+1)+4n(n+1)$

$=\dfrac{n(n+1)(n+4)(n+5)}{4}$

$\therefore \mathcal P=\displaystyle\sum_{n=1}^{\infty}\dfrac{\cancel{\color{#D61F06}{\dfrac{n(n+5)}2}}}{\dfrac{\cancel{\color{#D61F06}{n}}(n+1)(n+4)\cancel{\color{#D61F06}{(n+5)}}}{\cancelto2{\color{#D61F06}{4}}}}$

$=\displaystyle\sum_{n=1}^{\infty}\dfrac{2}{(n+1)(n+4)}=\dfrac 23\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{1}{n+1}-\dfrac{1}{n+4}\right)$

$\large (\color{#3D99F6}{ \mathbf{A~Telescopic~Series}})$

$\large =\dfrac 23\left(\dfrac 12+\dfrac 13+\dfrac 14\right)$

$=\large \dfrac{13}{18}$

$\large\therefore \sqrt{13+18+5}=\boxed{\color{#007fff}{6}}$

---

$(\color{#3D99F6}{**})$ When denominator can alternately be solved using Telescopic Series also.

$\dfrac 18\sum_{r=1}^n\left\{((r+6)-(r-2))(r+4)(r+2)(r)\right\}$

$= \dfrac 18\left[\sum_{r=1}^n\left\{(r+6)(r+4)(r+2)(r)\right\}-\left\{(r+4)(r+2)(r)(r-2)\right\}\right]$

$(\textbf{A Telescopic series})$

Which gives the same result as above.