How do you use the summation formulas to rewrite the expression #Sigma (4j+3)/n^2# as j=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
Nov 13, 2016

Answer:

# sum_(j=1)^n 1/n^2(4j+3) = (2n + 5)/n #

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Explanation:

Let # S_n = sum_(j=1)^n 1/n^2(4j+3) #
# :. S_n = 1/n^2sum_(j=1)^n (4j+3) #
# :. S_n = 1/n^2{sum_(j=1)^n (4j) + sum_(j=1)^n (3)} #
# :. S_n = 1/n^2{4 sum_(j=1)^n (j) + sum_(j=1)^n (3)} #

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

We have;

# S_n = 1/n^2{4 1/2n(n+1) + 3n} #
# :. S_n = 1/n^2(2n(n+1) + 3n) #
# :. S_n = 1/n^2(2n^2+2n + 3n) #
# :. S_n = 1/n^2(2n^2 + 5n) #
# :. S_n = 1/n^2(2n + 5)n #
# :. S_n = (2n + 5)/n #

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

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What happens as #n rarr oo#?

[ NB As an additional task we could possibly conclude that as #n rarr oo# then #S_n rarr 2#; This is probably the conclusion of this question]

Now, # S_n = (2n + 5)/n #

# :. S_n = 2n/n + 5/n #
# :. S_n = 2 + 5/n #

And so,

# lim_(n rarr oo) S_n = lim_(n rarr oo) (2 + 5/n) #
# :. lim_(n rarr oo) S_n = lim_(n rarr oo) (2) + lim_(n rarr oo)(5/n) #
# :. lim_(n rarr oo) S_n = 2 + 5 lim_(n rarr oo)(1/n) #
# :. lim_(n rarr oo) S_n = 2 #, #" as "(lim_(n rarr oo)(1/n) =0)#
Which confirms our assumption!