I used to wear contact lenses that were subject to protein deposits. 
So once a week, I would use an enzyme tablet that removed the protein
buildup.    I have no idea if the following numbers are accurate, but
it provides for an interesting thought exercise:  Let us say that in a
one week period, 100 "units" of protein accumulate on a lens.  Let us
also assume the enzyme is 99 percent effective in removing it, leaving
one unit of protein.  At the end of the second week, there are 101
units of protein, and again 99 percent is removed.    How much protein
is on the lens after 100 weeks?  1000 weeks?  More importantly, is the
amount convergent, or will the lens steadily have more and more
protein built up on it?

hi segwonk!!

Call k the amount of protein built up, and call n the number of the
week, we will name Wn the amount of protein remaining in a lens AFTER
clean it; so at the end of the week 1 after clean the lens we will
have:

W1 = Protein Built up - Protein Removed =
   = k - k * 0.99 = k * (1 - 0.99) = k * 0.01

Then at the end of the week 2 after clean the lens we will have:

W2 = Protein Built up - Protein Removed =
   = (k + W1) - (k + W1) * 0.99 = (k + W1) * 0.01 =
   = (k + k*0.01) * 0.01 = k * 0.01 + k * (0.01)^2


Then:
W3 = (k + W2) - (k + W2) * 0.99 = (k + W2) *0.01 =
   = (k + k * 0.01 + k * (0.01)^2) * 0.01 =
   = k * 0.01 + k * (0.01)^2 + k * (0.01)^3

In general:

Wn = k * 0.01 + k * (0.01)^2 + ... + k * (0.01)^n 
   = sum(from i=1 to n) [k * (0.01)^i] =
   = k * sum(from i=1 to n)[(0.01)^i] =
   = k * sum(from i=1 to n)[(1/100)^i]

Since:
1 + sum(from i=1 to n)[(1/100)^i] = sum(from i=0 to n)[(1/100)^i]
 
and 

sum(from i=0 to n)[(1/100)^i] = ((1 - (1/100)^n) / (1 - 1/100)) 

then:

Wn  = k * [((1 - (1/100)^n) / (1 - 1/100)) - 1]


Refered to the convergence we have that:

1 + sum(from i=1 to n)[(1/100)^i] = sum(from i=0 to n)[(1/100)^i]

When n tends to infinite this is a geometric series with ratio r =
1/100 < 1, then it converges to
1 / (1 - 1/100) = 1 / 99/100 = 100/99 = 1 + 1/99


See the following pages for references about the lasts calculations : 
"Geometric Series"
http://oregonstate.edu/dept/math/CalculusQuestStudyGuides/SandS/SeriesTests/geometric.html

"Dave's Math Tables: Geometric Summations":
http://www.math2.org/math/expansion/geom.htm 
 

Then if n tends to infinite:
sum(from i=1 to oo)[(1/100)^i] = sum(from i=0 to oo)[(1/100)^i] - 1
                               = 1/99

Then when n tends to infinite Wn tends to k/99 .


Note: If you want the amount before clean the lens at the end of the
week n (we will call this value as Pn) just use the following formula:
Pn = k + W(n-1) 


I hope this helps you. If you find something unclear or a point was
missed please let me know using the request for answer clarification
feature before rate this answer. I will be glad to give you further
assistance if it is needed.

Best regards.
livioflores-ga

Thanks.  Sorry took so long to rate -- I'm not great with math, so took awhile 
to understand it for myself.