How do you find a formula for the sum n terms #Sigma1/n^3(i-1)^2# and then find the limit as #n->oo#?

1 Answer
Nov 10, 2016

Answer:

# sum_(i=1)^n 1/n^3(i-1)^2 = 1/(6n^2)(2n-1)(n-1) #

# lim_(n rarr oo)sum_(i=1)^n 1/n^3(i-1)^2 = 1/3 #

Explanation:

Let # S_n = sum_(i=1)^n 1/n^3(i-1)^2 #
# :. S_n = 1/n^3 sum_(i=1)^n (i^2-2i+1) #
# :. S_n = 1/n^3 { sum_(i=1)^n i^2 - 2sum_(i=1)^ni + sum_(i=1)^n1 } #

And using the standard results:
# sum_(r=1)^n r = 1/2n(n+1) # and # sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) #

We have;

# S_n = 1/n^3 { 1/6n(n+1)(2n+1) -(2)1/2n(n+1) + n } #
# :. S_n = n/(6n^3) { (n+1)(2n+1) - 6(n+1) + 6 } #
# :. S_n = 1/(6n^2) { 2n^2+3n+1-6n-6+6 } #
# :. S_n = 1/(6n^2) { 2n^2-3n+1 } #
# :. S_n = 1/(6n^2)(2n-1)(n-1) #

Now we examine the behaviour of # S_n # as # n rarr oo #.
We have;

# S_n = 1/(6n^2) { 2n^2-3n+1 } #
# :. S_n = 1/3 -1/(2n)+1/(6n^2) #
# :. lim_(n rarr oo)S_n = lim_(n rarr oo) { 1/3 -1/(2n)+1/(6n^2) } #
# :. lim_(n rarr oo)S_n = lim_(n rarr oo) (1/3) - lim_(n rarr oo)(1/(2n)) + lim_(n rarr oo)(1/(6n^2)) #

And as both #1/n rarr 0# and #1/n^2 rarr 0# as #n rarr oo#, we have;

# :. lim_(n rarr oo)S_n = 1/3 - 0 + 0 #
# :. lim_(n rarr oo)S_n = 1/3 #