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

1 Answer
Jan 5, 2017

Answer:

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

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

Explanation:

Let # S_n = sum_(i=1)^n (1+i/n)(2/n) #
# :. S_n = sum_(i=1)^n (2/n+(2i)/n^2) #
# :. S_n = 2/n sum_(i=1)^n (1) + 2/n^2 sum_(i=1)^n (i)#

And using the standard results:

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

We have;

# S_n = 2/n(n)+2/n^2*1/2n(n+1) #
# :. S_n = 2+(n+1)/n #
# :. S_n = ((2n)+(n+1))/n #
# :. S_n = (3n+1)/n #

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

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

And as #1/n rarr 0# as #n rarr oo#, we have;

# :. lim_(n rarr oo)S_n = 3 #