Find the Limit?

#lim_(nrarroo)P_n#
where
#P_n=(2^3-1)/(2^3+1)*(3^3-1)/(3^3+1)*.....*(n^3-1)/(n^3+1); n=2,3,...#

1 Answer
Apr 7, 2018

Answer:

#lim_(n->oo) P_n =2/3#

Explanation:

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)#