Sum of Sequence of k x k!/Proof 2

Theorem

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then:

\(\ds \sum_{j \mathop = 1}^n j \times j!\)

\(=\)

\(\ds 1 \times 1! + 2 \times 2! + 3 \times 3! + \dotsb + n \times n!\)

\(\ds \)

\(=\)

\(\ds \paren {n + 1}! - 1\)

Proof

\(\ds \sum_{j \mathop = 1}^n j \times j!\)

\(=\)

\(\ds \sum_{j \mathop = 1}^n \paren {j + 1 - 1} \times j!\)

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 1}^n \paren {\paren {j + 1} j! - j!}\)

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 1}^n \paren {\paren {j + 1}! - j!}\)

\(\ds \)

\(=\)

\(\ds \paren {n + 1}! - 1!\)

Telescoping Series: Example 1

\(\ds \)

\(=\)

\(\ds \paren {n + 1}! - 1\)

$\blacksquare$