Question #1188e

#((1^2+2^2+3^2+...+n^2)(1^3+2^3+3^3+….+n^3))/(1^6+2^6+3^6+…+n^6)#
#=(n^5((1/n)^2+(2/n)^2+(3/n)^2+...+1^2)((1/n)^3+(2/n)^3+(3/n)^3+…+1^3))/(n^6((1/n)^6+(2/n)^6+(3/n)^6+…+1^6))#

#=(1/nsum_(i=0)^n(i/n)^2 1/nsum_(i=0)^n(i/n)^3)/(1/nsum_(i=0)^n(i/n)^6)#

As #n->oo#, using Riemann sums, we find
#=(int_0^1x^2\ dx\ int_0^1x^3\ dx)/(int_0^1x^6\ dx)#
#=([x^3/3]_0^1[x^4/4]_0^1)/[x^7/7]_0^1#
#=(1/3*1/4)/(1/7)#
#=7/12#

See the other solution I posted for an approach that does not resort to integration.

We know that #1^r+2^r+3^r+…+n^r# can be expressed as a formula in the form of a polynomial with degree #r+1#.

So, #1^2+2^2+3^2+…+n^2=an^3+bn^2+cn+d# for all #n# with certain #a,b,c,d#. This can be solved by setting #n=0,1,2,3# to get a system of linear equations:
#{(d=0),(a+b+c+d=1),(8a+4b+2c+d=5),(27a+9b+3c+d=14):}#

Solving this, we get #a=1/3,b=1/2,c=1/6,d=0#.

Thus, #1^2+2^2+3^2+…+n^2=1/3n^3+1/2n^2+1/6n=(2n^3+3n^2+n)/6#.

Use a similar process to find that #1^3+2^3+3^3+…+n^3=(n^4+2n^3+n^2)/4# and #1^6+2^6+3^6+…+n^6=(6n^7+21n^6+21n^5-7n^3+n)/42#.

Then, the original problem becomes
#lim_(n->oo)(((2n^3+3n^2+n)/6)((n^4+2n^3+n^2)/4))/((6n^7+21n^6+21n^5-7n^3+n)/42)#
#=lim_(n->oo)(42(2n^3+3n^2+n)(n^4+2n^3+n^2))/(24(6n^7+21n^6+21n^5-7n^3+n))#
#=7/12#.

See the other solution I posted for a method involving integration but is considerably easier.