Would you please tell me what's the limit of #(V_n-U_n)# as #n# approaches positive infinity ?
PS: #U_(n+1)= sqrt(U_nV_n)# and #V_(n+1)=(U_n+V_n)/2#

Question

Would you please tell me what's the limit of #(V_n-U_n)# as #n# approaches positive infinity ? 
PS: #U_(n+1)= sqrt(U_nV_n)# and #V_(n+1)=(U_n+V_n)/2#

Elijah Abram · Accepted Answer

#lim_(n->oo)(V_n-U_n) = 0# Making #U_n=lambda V_n# and #delta_n = V_n-U_n# we have #delta_(n+1)=V_(n+1)-U_(n+1)=(1-lambda)V_(n+1) = (lambda+1)/2V_n-sqrt(lambda)V_n# or #V_(n+1)/(V_n)= 1/2(lambda+1-2sqrt(lambda))/(1-lambda)=1/2(1-sqrt(lambda))^2/(1-(sqrt(lambda))^2)=1/2(1-sqrt(lambda))/(1+sqrt(lambda))# then #delta_(n+1) = V_0(1/2(1-sqrt(lambda))/(1+sqrt(lambda)))^n# so #lim_(n->oo)delta_n = 0# because #1/2((1-sqrt(lambda))/(1+sqrt(lambda)))<1#

Emily Ammerman · Answer

Here are my thoughts: # U_(n+1) =sqrt(U_nV_n) #
# V_(n+1) = 1/2(U_n+V_n) # Let # delta(n) = V_n - U_n # Assume the limit exists, and that:
# lim _(n rarr oo) (V_n - U_n) = epsilon # # :. lim _(n rarr oo) delta(n) = epsilon # Then it must also be the case that:
# :. lim _(n rarr oo) delta(n+1) = epsilon # # :. lim _(n rarr oo) (delta(n+1))/(delta(n)) = 1 # Now:
# delta(n+1) = V_(n+1) - U_(n+1) # # delta(n+1) = (U_n+V_n)/2 - sqrt(U_nV_n) # # (delta(n+1))/(delta(n)) = { (U_n+V_n)/2 - sqrt(U_nV_n) } / ( V_n - U_n ) = 1 # # (U_n+V_n)/2 - sqrt(U_nV_n)  = V_n - U_n #
# U_n+V_n - 2sqrt(U_nV_n)  = 2V_n - 2U_n #
# 3U_n-V_n = 2sqrt(U_nV_n)   #
# (3U_n-V_n)^2 = (2sqrt(U_nV_n))^2   #
# 9U_n^2 -6U_nV_n + V_n^2 = 4U_nV_n   #
# 9U_n^2 -10U_nV_n + V_n^2 = 0   #
# (9U_n - V_n)( U_n - V_n ) = 0   # # 9U_n - V_n = 0  => U_n = 1/9V_n #
# U_n - V_n  = 0 =>  U_n = V_n # I'm not sure if this is helpful, but feedback is welcome.

Isabella Ackerman · Answer

#lim n to oo (V_n-U_n)# is non-negative.

#U_(n+1)=sqrt(U_nV_n) to U_(n+1) > 0 to U_n > 0 to V_n > 0# Use the property of Means, #GM <= AM#. Here, #sqrt(U_nV_n) <=(U_n+V_n)/2# It follows that #U_(n+1)<=V_(n+1)#. So, #V_n-U_n>=0#, for n=2, 3, 4, ...# So, #lim n to oo (V_n-U_n)# is non-negative.

Ezra Adams · Answer

undefined The limit of (V_n-U_n) as n approaches positive infinity is 0.

Would you please tell me what's the limit of #(V_n-U_n)# as #n# approaches positive infinity ? PS: #U_(n+1)= sqrt(U_nV_n)# and #V_(n+1)=(U_n+V_n)/2#