The question is: For any integer, if n > m > 0
Evaluate 1/(m(m+1)) + 1/((m+1)(m+2)) + ... + 1/((n-1)n) + 1/(n(n+1))
I've gotten as far as rewriting it as (m-1)!/(m+1)! from m to n, but besides that I'm lost. Any help?
Evaluate 1/(m(m+1)) + 1/((m+1)(m+2)) + ... + 1/((n-1)n) + 1/(n(n+1))
I've gotten as far as rewriting it as (m-1)!/(m+1)! from m to n, but besides that I'm lost. Any help?
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Note that 1/(x(x+1)) = 1/x - 1/(x+1) by partial fractions.
Therefore,
1/(m(m+1)) + 1/((m+1)(m+2)) + ... + 1/((n-1)n) + 1/(n(n+1))
= (1/m - 1/(m+1)) + (1/(m+1) - 1/(m+2)) + ... + (1/(n-1) - 1/n) + (1/n - 1/(n+1))
= 1/m - 1/(n+1), since all other terms cancel in pairs.
I hope this helps!
Therefore,
1/(m(m+1)) + 1/((m+1)(m+2)) + ... + 1/((n-1)n) + 1/(n(n+1))
= (1/m - 1/(m+1)) + (1/(m+1) - 1/(m+2)) + ... + (1/(n-1) - 1/n) + (1/n - 1/(n+1))
= 1/m - 1/(n+1), since all other terms cancel in pairs.
I hope this helps!