How you solve this? #lim_(n->oo)(|__x__|+|__3^2x__|+...+|__(2n-1)^2x__|)/n^3#

#sum_(k=1)^n floor((2k)^2x)/n^3=sum_(k=1)^n floor(((2k)/n)^2x)(1/n)#

This can be understood as the realization of the Riemann-Stieltjes integral of

#int_0^2 y^2x dy = 8/3 x#

but

#sum_(k=1)^n floor(((2k)/n)^2x)(1/n) = sum_(k=1)^(n/2) floor(((2(2k-1))/n)^2x)(1/n)+sum_(k=1)^(n/2) floor(((2(2k))/n)^2x)(1/n)=2I_n#

Then #2I_n = 8/3 x# so #I_n = 4/3 x#

Let us first find a closed formula for:

#s_n = sum_(k=1)^n (2k-1)^2#

The first few terms are:

#color(blue)(1), 10, 35, 84, 165#

Write down the sequence of differences between consecutive terms:

#color(blue)(9), 25, 49, 81#

Write down the sequence of differences of those differences:

#color(blue)(16), 24, 32#

Write down the sequence of differences of those differences:

#color(blue)(8), 8#

Having arrived at a constant sequence (as expected after taking differences three times), we can use the initial term of each of these sequences to write down a formula for #s_n#:

#s_n = color(blue)(1)/(0!)+color(blue)(9)/(1!)(n-1)+color(blue)(16)/(2!)(n-1)(n-2)+color(blue)(8)/(3!)(n-1)(n-2)(n-3)#

#color(white)(s_n) = 1+9n-9+8n^2-24n+16+4/3n^3-8n^2+44/3n-8#

#color(white)(s_n) = 4/3n^3-1/3n#

Note also that:

#abs((floor(x)+floor(3^2x)+...+floor((2n-1)^2x))-(x+3^2x+...+(2n-1)^2x)) < n#

So:

#abs((floor(x)+floor(3^2x)+...+floor((2n-1)^2x))-(x+3^2x+...+(2n-1)^2x)) / n^3 < n/n^3 = 1/n^2 -> 0# as #n->oo#

So:

#lim_(n->oo) (floor(x)+floor(3^2x)+...+floor((2n-1)^2x))/n^3#

#= lim_(n->oo) (x+3^2x+...+(2n-1)^2x)/n^3#

#= lim_(n->oo) ((4/3n^3-1/3n)x)/n^3#

#= lim_(n->oo) ((4/3-1/(3n^2))x)#

#= 4/3x#