Sequence and series

Question

If tn= (n+2)(n+3)/4 for n=1,2,3.....

Then find the sum of 1/t1 + 1/t2 + 1/t3.... 1/t2003

Answer 1

1/t_n = 4/[(n+2)(n+3)].  Use partial fraction expansion:

 

4/[(n+2)(n+3)] = A/(n+2) + B/(n+3)

 

Multiply both sides by (n+2)(n+3):

 

4 = A(n+3) + B(n+2)

 

Solve for A by setting n = -2, and for B by setting n = -3.  The partial fraction expansion is then

 

1/t_n = 4/(n+2) - 4/(n+3).

 

Letting S_n be the sum of 1/t₁ + 1/t₂ + ... + 1/t_n, expansion produces

 

S_n = (4/3 - 4/4) + (4/4 - 4/5) + (4/5 - 4/6) + ... + [4/(n+2) - 4/(n+3)].

 

Notice that all terms except the first and last cancel, leaving

 

S_n = 4/3 - 4/(n+3).

 

So, S₂₀₀₃ = 4/3 - 4/2006.