How do you find a formula for the sum n terms #sum_(i=1)^n (16i)/n^3# and then find the limit as #n->oo#?

1 Answer
Jan 5, 2017

Answer:

# sum_(i=1)^n (16i)/n^3 = 18/n^2(n+1) #

# lim_(n rarr oo)sum_(i=1)^n (16i)/n^3 = 0 #

Explanation:

Let # S_n = sum_(i=1)^n (16i)/n^3 #
# :. S_n = 16/n^3 sum_(i=1)^n i #

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

We have;

# S_n = 16/n^3 { 1/2n(n+1) } #
# :. S_n = 18/n^2(n+1) #

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

# S_n = 18/n^2(n+1) #
# :. S_n = 18/n+18/n^2 #
# :. lim_(n rarr oo)S_n = lim_(n rarr oo) { 18/n+18/n^2 } #
# :. lim_(n rarr oo)S_n = 18lim_(n rarr oo) (1/n) + 18lim_(n rarr oo)(1/n^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 = 0 + 0 #
# :. lim_(n rarr oo)S_n = 0 #