Question

Evaluate (1/2!)+(2/3!)+(3/4!)+...(n/(n+1)!). Conjecture the sum if n tends to infinity.

Answer 1

Let the sum of the first n terms be Sn. Then we have:\[S_n =\sum_{k=1}^n\frac{k}{(k+1)!}=\sum_{k=1}^n\frac{(k+1)-1}{(k+1)!}=\sum_{k=1}^n\frac{k+1}{(k+1)!}-\frac{1}{(k+1)!}\]\[=\sum_{k=1}^n\frac{1}{k!}-\frac{1}{(k+1)!}\]This is a telescoping sum, so it should be clear what the summation equals. Now all thats left is to take the limit as n goes to infinity.

Answer 2

That's what I was going to say...