Lim n approaches infinity# 6/n((2n)/3 + (5n(n+1))/(2n) - (4n(n+1)(2n+1))/(6n^2))=#?

Question

lim n approaches infinity [6/n(2n/3 + 5n(n+1)/2n - 4n(n+1)(2n+1)/6n^2)]=?

Emily Ammerman · Accepted Answer

# lim_(n rarr oo) 6/n[(2n)/3 +( 5n(n+1))/(2n) - (4n(n+1)(2n+1))/(6n^2) ] = 11 #

Let # S = 6/n[(2n)/3 +( 5n(n+1))/(2n) - (4n(n+1)(2n+1))/(6n^2) ] # # :. S = 6[ 1/(6n^2){ 4n^2 + 15n(n+1) - 4(n+1)(2n+1) } ] # # :. S = 1/(n^2)[ 4n^2 + 15n^2+15n - 4(2n^2+3n+1) ] # # :. S = 1/(n^2)[ 4n^2 + 15n^2+15n - 8n^2-12n-4 ] # # :. S = ( 11n^2 + 3n - 4 ) / (n^2)# # :. S = 11 + 3/n - 4/n^2 # So: # lim_(n rarr oo) S = lim_(n rarr oo) (11 + 3/n - 4/n^2) # # lim_(n rarr oo) S = lim_(n rarr oo) 11 + 3lim_(n rarr oo) 1/n - 4lim_(n rarr oo)1/n^2 # # lim_(n rarr oo) S = 11 + 0 + 0# # lim_(n rarr oo) S = 11 # This ca be verified by looking at the graph of #y=11 + 3/x - 4/x^2# graph{11 + 3/x - 4/x^2 [-27.95, 76.05, -19, 33]}

Charles Anctil · Answer

I'm guessing this is typed incorrectly.

The term 5n(n+1)/2n means #5n(n+1)/2n# whose limit is #oo#. My guess is the question wants (5n(n+1))/(2n) which is #(5n(n+1))/(2n) # Similarly, 4n(n+1)(2n+1)/6n^2 is #4n(n+1)(2n+1)/6n^2 # which I suppose should be #(4n(n+1)(2n+1))/(6n^2) # So here's what I guess the expression should be: #6/n((2n)/3+(5n(n+1))/(2n)-(4n(n+1)(2n+1))/(6n^2))# # = (4n)/n + (15n(n+1))/n^2 - (4n(n+1)(2n+1))/n^3# # = 4 +15((n/n)((n+1)/n)) - 8(n/n)((n+1)/n)((2n+1)/n)# The limit as #n rarroo# is #4 +15(1)(1)-8(1)(1)(2) = 4+15-8 = 11#

Emily Ammerman · Answer

#11# After expanding #f(n)= 6/n((2n)/3 + (5n(n+1))/(2n) - (4n(n+1)(2n+1))/(6n^2))=11+3/n-4/n^2# so #lim_(n->oo)f(n)=lim_(n->oo)(11+3/n-4/n^2)=11#

Ezra Adams · Answer

undefined The limit of the given expression as $n$ approaches infinity is $\frac{1}{3}$.