Quickie #6

Find the sum of the series:

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

Quote:

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Originally Posted by Soroban

Find the sum of the series:

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

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Quick already solved it.
(n+1)!-1

It is just mathematical iduction

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.