Show that `1/(j*(j+1)) = (1/j) - (1/(j - 1))` and evaluate `sum_(j=1)^n 1/(j*(j+1)) ` .

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We have to evaluate `sum_(j = 1)^n (1/(j*(j+1)))`

Let `1/(j*(j +1))` = `a/j + b/(j + 1)`

=> `1/(j*(j+1))` = `(a*(j+1) + b*j)/(j*(j+1))`

=> `a*j + a + b*j = 1`

=> `a + b = 0` and a = 1

=> b = -1 and a = 1

This gives `1/(j*(j+1)) = 1/j - 1/(j +1)`

The sum `sum_(j = 1)^n 1/(j*(j+1)) = sum_(j=1)^n 1/j + 1/(j +1)`

=> `sum_(j = 1)^n (1/j) + sum_(j = 1)^(n) 1/(j+1)`

The sum of the reciprocals of the first n numbers `1/1 + 1/2 + 1/3+ ... +1/n` is a diverging series. The value of the sum does not does not have a finite value.

`sum_(j=1)^n 1/(j*(j+1))` does not have a finite value and there is no general expression to calculate it.

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