Evaluate the sum #sum_(i=1)^n (12i^2(i-1))/n^4# for #n=10,100,1000# and #10000#?

1 Answer
Jul 17, 2017

Answer:

# {: (n=10, => "sum"=3.168), (n=100, => "sum"=3.019698), (n=1000, => "sum"=3.001996998), (n=10000, => "sum"=3.00019997) :} #

Explanation:

Let:

# S_n = sum_(i=1)^n (12i^2(i-1))/n^4 #
# " " = 12/n^4sum_(i=1)^n {i^3-i^2} #
# " " = 12/n^4{sum_(i=1)^n i^3 - sum_(i=1)^n i^2} #

And using the standard results:

# sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) #
# sum_(r=1)^n r^3 = 1/4n^2(n+1)^2 #

We have;

# S_n = 12/n^4{1/4n^2(n+1)^2 - 1/6n(n+1)(2n+1) }#
# " " = 12/n^4{1/4n^2(n+1)^2 - 1/6n(n+1)(2n+1) }#
# " " = 1/n^3{3n(n+1)^2 - 2(n+1)(2n+1) }#
# " " = (n+1)/n^3{3n(n+1) - 2(2n+1) }#
# " " = (n+1)/n^3{3n^2+3n - 4n - 2 }#
# " " = (n+1)/n^3{3n^2- n - 2 }#
# " " = {(n+1)(n-1)(3n+2) }/n^3#

And this has been calculated using Excel for #n=10, 100, 1000, 10000#

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Conclusion: What happens as #n rarr oo#?

[ NB As an additional task we could possibly conclude that as #n rarr oo# then #S_n rarr 1#; This is probably the conclusion of this question]

From the above results it looks as if:

# S_n ~~ 3# as #n rarr oo#

Let us see if this is actually the case. We can manipulate #S_n# as follows;

# S_n = ( (n+1)(3n^2- n - 2)) / n^3 } #
# " " = ( 3n^3- n^2 - 2n + 3n^2- n - 2) / n^3 #
# " " = ( 3n^3+ 2n^2 - 3n - 2) / n^3 #
# " " = 3+ 2/n - 3/n^2 - 2/n^3 #

And so,

# lim_(n rarr oo) S_n = lim_(n rarr oo) (3+ 2/n - 3/n^2 - 2/n^3) #
# " " = 3+ 0 - 0 - 0 #
# " " = 3 #

Which confirms our numerical observations!