We see that
#P_n = prod_(k=2)^n (k^3-1)/(k^3+1) = prod_(k=2)^n(k^3-1^3)/(k^3+1^3)#
Both the numerator and denominator are a sum or difference of cubes, for which we have a formula:
#a^3+b^3 = (a+b)(a^2-ab+b^2)#
#a^3-b^3=(a-b)(a^2+ab+b^2)#
Thus, we have
#P_n = prod_(k=2)^n ((k-1)(k^2+k+1))/((k+1)(k^2-k+1))#
#P_n = ((2-1)(2^2+2+1))/((2+1)(2^2-2+1))*((3-1)(3^2+3+1))/((3+1)(3^2-3+1))...((n-2)(n^2-n+1))/((n)(n^2-3n+3))((n-1)(n^2+n+1))/((n+1)(n^2-n+1))#
We can group #P_n# into the following products:
#P_n = color(red)(prod_(k=2)^n (k-1)/(k+1))*((2^2+2+1)/(2^2-2+1))((3^2+3+1)/(3^2-3+1))...((n^2-n+1)/(n^2-3n+3))((n^2+n+1)/(n^2-n+1))#
Let the product in #color(red)("red")# be #p_n#.
Notice how some things cancel each other:
#P_n = color(red)(p_n)*(cancel(2^2+2+1)/(2^2-2+1))(cancel(3^2+3+1)/cancel(3^2-3+1))...(cancel(n^2-n+1)/cancel(n^2-3n+3))((n^2+n+1)/cancel(n^2-n+1))#
It might seem confusing, but the numerator of the #k#-th term is being cancelled by the denominator of the #k+1#-th term.
#:.#
#P_n = p_n (n^2+n+1)/3#
Let's write out #p_n#:
#p_n = (2-1)/(2+1) * (3-1)/(3+1)...(n-2)/(n) * (n-1)/(n+1)#
#p_n = (1*2*3*...*(n-2)(n-1))/(3*4*5*...*n(n+1))#
Again, lots of things cancel out.
#p_n = (1*2*cancel(3*4*5*...*(n-2)(n-1)))/(cancel(3*4*5*...*(n-2)(n-1)) n(n+1)) = 2/(n(n+1))#
Therefore, we have:
#P_n = color(red)(2/(n(n+1))) *[(n(n+1))/3 + 1/3]#
#P_n = 2/3[1+1/(n(n+1))]#
Now, let's take the #color(red)("limit")# as #n->oo# of both sides.
#lim_(n->oo) P_n = lim_(n->oo) 2/3 [1+1/(n(n+1))]#
As #n# grows boundless, #1/(n(n+1))# becomes arbitrarily small.
In other words, it approaches #0#.
#=> lim_(n->oo) P_n = 2/3 [1+0] = 2/3#
#color(red)( lim_(n->oo) P_n = 2/3)#
