The answer is: 12.
First of all: "How much is the sum of a geometric series?".
#sum_(n=0)^(+oo)r^n=a_0/(1-r)#, where r (such as #|r|<1#) is called "ratio" (#r=a_k/a_(k-1)#) and #a_0# is the first term of the series (in this case #a_0=r^0=1#).
Let's list all the terms of the series:
#1*3/4 + 2*(3/4)^2+3*(3/4)^3+4*(3/4)^4+...#, but we can also say:
#3/4+9/16+9/16+27/64+27/64+27/64+81/256+81/256+81/256+81/256+...#,
or, better:
#3/4+9/16+27/64+81/256+...+#
#9/16+27/64+81/256+...+#
#27/64+81/256+...+#
#81/256+...+#
#...#
The first sum is:
#sum_(n=1)^(+oo)(3/4)^n#,
The second sum is:
#sum_(n=2)^(+oo)(3/4)^n#,
The third sum is:
#sum_(n=3)^(+oo)(3/4)^n#,
The fourth sum is:
#sum_(n=4)^(+oo)(3/4)^n#,
#...#
So:
#sum_(n=1)^(+oo)n(3/4)^n=sum_(n=1)^(+oo)(3/4)^n+sum_(n=2)^(+oo)(3/4)^n+ sum_(n=3)^(+oo)(3/4)^n+sum_(n=4)^(+oo)(3/4)^n+...=#.
#=3/4*1/(1-3/4)+9/16*1/(1-3/4)+27/64*1/(1-3/4)+81/256*1/(1-3/4)+...=#
#=3/4*4+9/16*4+27/64*4+81/256*4+...=#
#=4(3/4+9/16+27/64+81/256+...)=#
#=4*sum_(n=1)^(+oo)(3/4)^n=4*3/4*1/(1-3/4)=4*3/4*4=12#.
