Let's express this as a sum:
#sum_(k=1)^oo 500(1.12)^-k#
Since #1.12=112/100=28/25#, this is equivalent to:
#sum_(k=1)^oo 500(28/25)^-k#
Using the fact that #(a/b)^-c=(1/(a/b))^c=(b/a)^c#, we have:
#sum_(k=1)^oo 500(25/28)^k#
Also, we can pull the #500# out of the summation sign, like this:
#500sum_(k=1)^oo (25/28)^k#
Alright, now what is this? Well, #sum_(k=1)^oo(25/28)^k# is what's known as a geometric series. Geometric series involve an exponent, which is exactly what we have here. The awesome thing about geometric series like this one is that they sum up to #r/(1-r)#, where #r# is the common ratio; i.e. the number that's raised to the exponent. In this case, #r# is #25/28#, because #25/28# is what's raised to the exponent. (Side note: #r# has to be between #-1# and #1#, or else the series doesn't add up to anything.)
Therefore, the sum of this series is:
#(25/28)/(1-25/28)#
#=(25/28)/(3/28)#
#=25/28*28/3=25/3#
We've just discovered that #sum_(k=1)^oo (25/28)^k=25/3#, so the only thing that's left is to multiply it by #500#:
#500sum_(k=1)^oo (25/28)^k#
#=500*25/3#
#=12500/3~~4166.667#
You can find out more about geometric series here (I encourage you to watch the whole series Khan Academy has on geometric series).
