Find the sum of the series:

. . $\displaystyle 1(1!) + 2(2!) + 3(3!) + \cdots + n(n!)$

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- Dec 30th 2006, 01:02 PMSorobanQuickie #6
Find the sum of the series:

. . $\displaystyle 1(1!) + 2(2!) + 3(3!) + \cdots + n(n!)$

- Dec 30th 2006, 01:43 PMThePerfectHacker
- Dec 30th 2006, 02:41 PMTD!
I don't know where it was solved, but even without (explicit) induction:

$\displaystyle

k\left( {k!} \right) = \left( {k + 1} \right)k! - k! = \left( {k + 1} \right)! - k!

$

So summing for k = 1 till n gives (n+1)!-1 since all intermediate terms cancel out.