How do you use the summation formulas to rewrite the expression #Sigma (2i+1)/n^2# as i=1 to n without the summation notation and then use the result to find the sum for n=10, 100, 1000, and 10000?

1 Answer
Dec 10, 2016

Answer:

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

enter image source here

Explanation:

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

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

We have;

# \ \ \ \ \ S_n = 1/n^2{2*1/2n(n+1) + n} #
# :. S_n = 1/n^2{n(n+1) + n} #
# :. S_n = 1/n^2{n^2+n + n} #
# :. S_n = 1/n^2{n^2+2n} #
# :. S_n = 1/n^2{n(n+2)} #
# :. S_n = (n+2)/n #

And this has been calculated using Excel for #n=10, 100, 1000, 10000#

enter image source here

What happens as #n rarr oo#?

[ NB As an additional task we could possibly conclude that as #n rarr oo# then #S_n rarr 1#; This is probably the conclusion of this question]

Now, # S_n = (n+2)/n #

# :. S_n = 1+2/n #

And so,

# lim_(n rarr oo) S_n = lim_(n rarr oo) (1+2/n) #
# :. lim_(n rarr oo) S_n = lim_(n rarr oo) (1) + 2lim_(n rarr oo)(1/n) #
# :. lim_(n rarr oo) S_n = 1+0 #
# :. lim_(n rarr oo) S_n = 1#
Which confirms our assumption!