Find a general formula for S(n)=sum_4^(n)(1/(k-3)-1/(k))S(n)=n∑4(1k−3−1k) and evaluate the limit S=lim_(n rarr oo) S(n)?
1 Answer
sum_4^(n)(1/(k-3)-1/(k)) = ( (n-3)(11n^2-18n+4) ) / ( 6n(n-1)(n-2) )
sum_4^(oo)(1/(k-3)-1/(k)) = 11/6
Explanation:
We seek:
S = sum_4^(oo)(1/(k-3)-1/(k))
Firstly, let us find a formula for:
S(n) = sum_4^(n)(1/(k-3)-1/(k))
This a a typical summation of differences, and as such many terms will cancel. If we expand the summation this becomes clear:
S(n) = (1/1 - color(blue)(1/4)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=4)
\ \ \ \ \ \ \ \ \ \ \ \ \ (1/2 - color(green)(1/5)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=5)
\ \ \ \ \ \ \ \ \ \ \ \ \ (1/3 - color(red)(1/6)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=6)
\ \ \ \ \ \ \ \ \ \ \ \ \ (color(blue)(1/4) - 1/7) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=7)
\ \ \ \ \ \ \ \ \ \ \ \ \ (color(green)(1/5) - 1/8) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=8)
\ \ \ \ \ \ \ \ \ \ \ \ \ (color(red)(1/6) - 1/9) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=9)
\ \ \ \ \ \ \ \ \ \ \ \ \ vdots
\ \ \ \ \ \ \ \ \ \ \ \ \ (1/(n-6)-color(purple)(1/(n-3))) + \ \ \ \ \ \ (k=n-3)
\ \ \ \ \ \ \ \ \ \ \ \ \ (1/(n-5)-1/(n-2)) + \ \ \ \ \ \ (k=n-2)
\ \ \ \ \ \ \ \ \ \ \ \ \ (1/(n-4)-1/(n-1)) + \ \ \ \ \ \ (k=n-1)
\ \ \ \ \ \ \ \ \ \ \ \ \ (color(purple)(1/(n-3))-1/(n)) + \ \ \ \ \ \ \ \ \ \ \ \ \ (k=n)
And now if we cancel the terms which appear as positive from one term and negative and another:
S(n) = (1/1 - color(blue)(cancel(1/4))) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=4)
\ \ \ \ \ \ \ \ \ \ \ \ \ (1/2 - color(green)(cancel(1/5))) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=5)
\ \ \ \ \ \ \ \ \ \ \ \ \ (1/3 - color(red)(cancel(1/6))) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=6)
\ \ \ \ \ \ \ \ \ \ \ \ \ (color(blue)(cancel(1/4)) - cancel(1/7)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=7)
\ \ \ \ \ \ \ \ \ \ \ \ \ (color(green)(cancel(1/5)) - cancel(1/8)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=8)
\ \ \ \ \ \ \ \ \ \ \ \ \ (color(red)(cancel(1/6)) - cancel(1/9)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=9)
\ \ \ \ \ \ \ \ \ \ \ \ \ vdots
\ \ \ \ \ \ \ \ \ \ \ \ \ (cancel(1/(n-6))-color(purple)(cancel(1/(n-3)))) + \ \ \ \ \ \ (k=n-3)
\ \ \ \ \ \ \ \ \ \ \ \ \ (cancel(1/(n-5))-1/(n-2)) + \ \ \ \ \ \ (k=n-2)
\ \ \ \ \ \ \ \ \ \ \ \ \ (cancel(1/(n-4))-1/(n-1)) + \ \ \ \ \ \ (k=n-1)
\ \ \ \ \ \ \ \ \ \ \ \ \ (color(purple)(cancel(1/(n-3)))-1/(n)) + \ \ \ \ \ \ \ \ \ \ \ \ \ (k=n)
Which leaves us with a much reduced collection of terms:
S(n) = 1/1 + 1/2 +1/3 -1/(n-2) -1/(n-1) -1/n
\ \ \ \ \ \ \ \ = 11/6 -1/(n-2) -1/(n-1) -1/n ..... [A]
Although not strictly required, We can simplify this result (this is typically the case in an exam question when
\ \ = 11/6 -1/(n-2) -1/(n-1) -1/n
\ \ = ( 11n(n-1)(n-2) - 6n(n-1) -6n(n-2) - 6(n-1)(n-2) ) / ( 6n(n-1)(n-2) )
\ \ = ( 11n(n-1)(n-2) - 6n(n-1) -6n(n-2) - 6(n-1)(n-2) ) / ( 6n(n-1)(n-2) )
\ \ = ( 11n^3-33n^2+22n -6n^2+6n -6n^2+12n -6n^2+18n-12 ) / ( 6n(n-1)(n-2) )
\ \ = ( 11n^3-51n^2+58n-12 ) / ( 6n(n-1)(n-2) )
\ \ = ( (n-3)(11n^2-18n+4) ) / ( 6n(n-1)(n-2) )
Factorization was performed via a calculator at the last step, the answer at [A] would suffice
So finally, we seek:
S = lim_(n rarr oo) S(n)
\ \ = lim_(n rarr oo) {11/6 -1/(n-2) -1/(n-1) -1/n }\ \ \ \ \ \ \ (from [A])
\ \ = 11/6
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Validation:
It is always worth checking a few sums just to check the result is consistent
By direct summation
S(4) = 3/4
S(5) = 21/21
S(5) = 73/60
And using the derived summation formula gives the same result
