Crystal Shield/Proof of the simplification

We start from the main equation:

$$ r_{avg} = \frac{ \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n}p \left[ \sum\limits_{i=1}^{n} d^{(0)}_{i,n} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} \left(d^{(1)}_{i,n} - \left\lfloor{.05 \times R \times d^{(1)}_{i,n}}\right\rfloor\right) +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil} d^{(2)}_{i,n} \right] }{           \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n}p \left[ \sum\limits_{i=1}^{n} d^{(0)}_{i,n} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} d^{(1)}_{i,n} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil} d^{(2)}_{i,n} \right] } $$

The second summation within the brackets in de numerator can be split up.

$$ r_{avg} = \frac{ \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n}p \left[ \sum\limits_{i=1}^{n} d^{(0)}_{i,n} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} d^{(1)}_{i,n} -                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} \left\lfloor{.05 \times R \times d^{(1)}_{i,n}}\right\rfloor +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil} d^{(2)}_{i,n} \right] }{           \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n}p \left[ \sum\limits_{i=1}^{n} d^{(0)}_{i,n} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} d^{(1)}_{i,n} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil} d^{(2)}_{i,n} \right] } $$

Split up the damage reduction component.

$$ r_{avg} = \frac{ \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n}p \left[ \sum\limits_{i=1}^{n} d^{(0)}_{i,n} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} d^{(1)}_{i,n} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil} d^{(2)}_{i,n} \right] -           \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n}p \left[ \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} \left\lfloor{.05 \times R \times d^{(1)}_{i,n}}\right\rfloor \right] }{           \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n}p \left[ \sum\limits_{i=1}^{n} d^{(0)}_{i,n} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} d^{(1)}_{i,n} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil} d^{(2)}_{i,n} \right] } $$

The first term of the numerator is equal to the denominator.

$$ r_{avg} = 1 -   \frac{ \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n}p \left[ \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} \left\lfloor{.05 \times R \times d^{(1)}_{i,n}}\right\rfloor \right] }{           \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n}p \left[ \sum\limits_{i=1}^{n} d^{(0)}_{i,n} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} d^{(1)}_{i,n} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil} d^{(2)}_{i,n} \right] } $$

If we want to know the damage reduction over a large number of hits, we could use the statistical average of the received damage, this means $$d^{(j)}_{i,n}$$ becomes a constant value $$\bar{d}$$. We also drop the floor function in the numerator and add a value $$\phi \in [0,1)$$ to correct for the error.

$$ d^{(j)}_{i,n} \Rightarrow \bar{d} $$

$$ \Rightarrow r_{avg} = 1 -   \frac{ \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n}p \left[ \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} (.05 \times R \times \bar{d} - \phi) \right] }{           \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n}p \left[ \sum\limits_{i=1}^{n} \bar{d} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} \bar{d} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil} \bar{d} \right] } $$

The second summation in the numerator has no terms that change on increasing $$i$$, the summation can thus be simplified to a product of the range of the index and the summands.

$$ r_{avg} = 1 -   \frac{ \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n}p \left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil \left(.05 \times R \times \bar{d} - \phi\right) }{           \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n}p \left[ \sum\limits_{i=1}^{n} \bar{d} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} \bar{d} +                   \sum\limits_{i=1}^{\left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil} \bar{d} \right] } $$

Place terms not dependent on $$n$$ outside of the summations.

$$ \begin{aligned} r_{avg} &= 1 -   \frac{ p \left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil \left(.05 \times R \times \bar{d} - \phi\right) \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n} }{           p \; \bar{d} \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n} \left[ \sum\limits_{i=1}^{n} 1                   +                    \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} 1                   +                    \sum\limits_{i=1}^{\left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil} 1               \right] } \\ & =    1 -    \frac{ p \; \bar{d} \left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil \left(.05 \times R - \frac{\phi}{\bar{d}}\right) \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n} }{           p \; \bar{d} \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n} \left[ \sum\limits_{i=1}^{n} 1                   +                    \sum\limits_{i=1}^{\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil} 1                   +                    \sum\limits_{i=1}^{\left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil} 1               \right] }\end{aligned} $$

Drop common term in numerator and denominator.

$$ r_{avg} = 1 -   \frac{ \left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil \left(.05 \times R - \frac{\phi}{\bar{d}}\right) \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n} }{           \sum\limits_{n=0}^{\infty} \left(1-p\right)^{n} \left[ n + \left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil + \left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil \right] } \tag{1} $$

To simplify even further we need to find the infinite sum equivalents of the remaining summations. Since $$(1-p)=x<1$$ (p = perk proc chance), the following is true

$$ \tag{2} \sum\limits_{n=0}^{\infty} ax^{n} = \frac{a}{1-x} $$

Take the derivative of (2):

$$ \begin{aligned} \sum\limits_{n=0}^{\infty} anx^{n-1} &= \frac{a}{(1-x)^2} \\ \sum\limits_{n=0}^{\infty} a(n+1)x^{n} &= \frac{a}{(1-x)^2} \\ \sum\limits_{n=0}^{\infty} anx^{n} + \sum\limits_{n=0}^{\infty} ax^{n} &= \frac{a}{(1-x)^2} \tag{3}\end{aligned} $$

The second term in (3) is equal to (2):

$$ \begin{aligned} \sum\limits_{n=0}^{\infty} anx^{n} &= \frac{a}{(1-x)^2} - \frac{a}{1-x} \\ &= \frac{ax}{(1-x)^2}\end{aligned} $$

Combine (2) and (3) with diffent constants:

$$ \begin{aligned} \sum\limits_{n=0}^{\infty} anx^{n} + \sum\limits_{n=0}^{\infty} bx^{n} &= \frac{ax}{(1-x)^2} + \frac{b}{1-x}\\ \sum\limits_{n=0}^{\infty} x^{n}(an+b) &= \frac{ax + b(1-x)}{(1-x)^2} \tag{4}\end{aligned} $$

If we take $$ x=(1-p) $$ and $$ a=1 $$ in (2) we get:

$$ \sum\limits_{n=0}^{\infty} (1-p)^{n} = \frac{1}{p} \tag{5} $$

For $$x=(1-p)$$, $$a=1$$ and $$b=\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil + \left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil$$ in (4), we get:

$$ \sum\limits_{n=0}^{\infty} (1-p)^{n} \left[ n + \left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil + \left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil \right] = \frac{(1-p) + \left(\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil + \left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil\right)p}{p^2} \tag{6} $$

The summations (5) and (6) occur in (1) and can be substituted:

$$ \begin{aligned} \Rightarrow r_{avg} &= 1 -   \frac{ \left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil \left(.05 \times R - \frac{\phi}{\bar{d}}\right) \frac{1}{p} }{           \frac{ p \left(\left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil +                    \left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil\right) - p + 1 }{                   p^2 }       } \\ & =     1 -    \frac{ \left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil \left(.05 \times R - \frac{\phi}{\bar{d}}\right) }{           \left\lceil{\frac{t_{abs}}{t_{AS}}}\right\rceil + \left\lceil{\frac{t_{cd}-t_{abs}}{t_{AS}}}\right\rceil - 1 + \frac{1}{p} }\end{aligned} $$

Where $$\phi = .05 \times R \times \bar{d} - \left\lfloor{.05 \times R \times \bar{d}}\right\rfloor$$. This represents the rounding error introduced by dropping the floor function. $$\phi \in [0,1)$$, so for $$\bar{d} > 500$$ we have <4% error for ignoring $$\phi$$ and can therefor safely be ignored for most practical situations.