How do you calculate this?

Question

¹⁄₂₍₄₎ + ¹⁄₄₍₆₎ + ... + ¹⁄₂₀₁₄₍₂₀₁₆₎ + ¹⁄₂₀₁₆₍₂₀₁₈₎ = ?

Answer 1

the general term is:

1/ [(2n)(2n+2)]

for n=1,2,3,4......,1008

1/ [(2n)(2n+2)] = 1/ [4 n(n+1)]

(1/4) (1/(n(n+1))

partial fraction decomposition on 1/(n(n+1)) = a/n + b/(n+1)

1 = a(n+1) + bn

= an + a + bn

= (a+b)n+a

a+b=0 and a=1

so b = -1

the partial fraction decomposition for the sequence is 1/n - 1/(n+1)

Sk = SUM [ 1/n - 1/(n+1)], n=1,2,3,....,k

= SUM [1/n], n=1,2,3,...,k - SUM [ 1/(n+1)], n=1,2,3,....,k

= SUM [1/n], n=1,2,3,...,k - SUM [ 1/n], n=2,3,4,....,k+1

= 1 + SUM [1/n], n=2,3,...,k - SUM [1/n], n=2,3,4,....k - 1/(k+1) =

1 - 1/(k+1)

So for k=1008, the sum is

1 - 1/1009 = (1009-1)/1009 = 1008/1009