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Let S = {n: 4 to 2017) Σ[1/(n^2 – 5n + 6)]

Let T = 1/(n^2 – 5n + 6) = 1/(n – 3) – 1/(n – 2)

n = 4, T = 1 – 1/2

n = 5, T = 1/2 – 1/3

n = 6, T = 1/3 – 1/4

Intermediate terms cancel, (telescope)

Sum of 4th to term t is 1 – 1/(t – 2),

S = 1 – 1/(2015) = 2014/2015

Decompose the fraction

n^2 - 5n + 6 = (n - 3) * (n - 2)

1 / (n^2 - 5n + 6) = a / (n - 3) + b / (n - 2)

1 = a * (n - 2) + b * (n - 3)

0n + 1 = an + bn - 2a - 3b

0n = an + bn

0 = a + b

a = -b

1 = -2a - 3b

-1 = 2a + 3b

-1 = -2b + 3b

-1 = b

a = -b

a = 1

1 / (n^2 - 5n + 6) = a / (n - 3) + b / (n - 2)

1 / (n^2 - 5n + 6) = 1 / (n - 3) - 1 / (n - 2)

Now we have:

1 / (4 - 3) - 1 / (4 - 2) + 1 / (5 - 3) - 1 / (5 - 2) + 1 / (6 - 3) - 1 / (6 - 2) + 1 / (7 - 3) - 1 / (7 - 2) + .... + 1 / (2017 - 3) - 1 / (2017 - 2) =>

1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + .... - 1/2014 + 1/2014 - 1/2015 =>

1 + 0 + 0 + 0 + ... + 0 - 1/2015 =>

2015/2015 - 1/2015 =>

(2015 - 1) / 2015 =>

2014 / 2015

Sigma(from n = 4 to 2017) 1/(n^2 - 5n + 6)

= 2014/2015

Decimal approximation:

0.999503722084367245657568238213399503722084367245657568238...