Question

In this problem we evaluate the series:

1/(1x3)+1/(2x4)+1/(3x5)...+1/(98x100)

A. Notice that each fraction in the sum has the form 1/(n(n+2)) for some positive integer n. Find constants A and B such that 1/(n(n=2)=A//n+B/(n+2)

B. Use your answer to part A to find the desired sum

Answer 1

A. 1/(n(n+2)) = (1/2)/n - (1/2)/(n+2)

B. 14651/19800

Explanation:

A.

The coefficients of the partial-fraction expansion can be found from ...

f(n) = 1/(n(n+2)) = A/n +B/(n+2)

n·f(x) = 1/(n+2) = A +Bn/(n+2)

For n=0, this becomes ...

1/(0 +2) = A = 1/2

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Similarly, ...

(n+2)·f(n) = 1/n = A(n+2)/n +B

For n = -2, this becomes ...

1/(-2) = B = -1/2

The n-th term can be written ...

1/(n(n+2)) = (1/2)/n - (1/2)/(n+2)

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B.

The sum is ...

1/(1·3) +1/(2·4) +1/(3·5) +... +1/(98·100)

= 1/2(1/1 -1/3 +1/2 -1/4 +1/3 -1/5 +... +1/98 -1/100)

= 1/2((1/1 +1/2 +1/3 +...1/98) -(1/3 +1/4 +1/5 +...+1/100)

We notice that terms 3..98 cancel, so the sum is ...

= 1/2(1/1 +1/2 -1/99 -1/100) = (1/2)(3/2 -199/9900)

= 14651/19800