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

1 Answer
Nov 20, 2016

Answer:

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

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

Explanation:

We will need the standard results:
# sum_(r=1)^n r = 1/2n(n+1) #

Let

# S_n = sum_(i=1)^n (2i)/n(2/n) #
# :. S_n = 4/n^2sum_(i=1)^n i #
# :. S_n = 4/n^2{ 1/2n(n+1) } #
# :. S_n = 2/n(n+1) #
# :. S_n = 2 + 2/n #

Hence, # sum_(i=1)^n (2i)/n(2/n) = 2 + 2/n #

Consequently,

# lim_(n rarr oo) S_n = lim_(n rarr oo) (2 + 2/n) #
# lim_(n rarr oo) S_n = lim_(n rarr oo) (2) + lim_(n rarr oo) (2/n) #
# lim_(n rarr oo) S_n = 2 + 0 #
# lim_(n rarr oo) S_n = 2 #