How do I prove that $$\sum\limits_{r=1}^{n} r(r!) = (n+1)!-1$$
I was able to get to factor: $LHS = k(k!) + (k+1)(k+1)!$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\, RHS = (k+2)! -1$
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Have a look around the site --- I'm sure this question has been asked before. – Gerry Myerson Nov 02 '13 at 23:49
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $$ \color{#0000ff}{\large\sum_{r = 1}^{n}r\pars{r!}} = \sum_{r = 1}^{n}\bracks{\pars{r + 1}\pars{r!} - r!} = \sum_{r = 1}^{n}\bracks{\pars{r + 1}! - r!} $$ $$ \color{#0000ff}{\large\sum_{r = 1}^{n}r\pars{r!}} = \pars{2! - 1!} + \pars{3! - 2!} + \cdots + \bracks{\pars{n + 1}! - n!} = \pars{n + 1}! - 1! = \color{#0000ff}{\large\pars{n + 1}! - 1} $$
Felix Marin
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Is this an improvement on the answers in the duplicate question that I pointed out? If so, that's where this answer belongs, so it's all together in one place. – Gerry Myerson Nov 03 '13 at 00:22
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@GerryMyerson I didn't know there was a previous answer before you mentioned it. I just checked that and I could see that one is based in the Induction Principle. Thanks for call that to my attention. – Felix Marin Nov 03 '13 at 00:33
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Did you read the comments before you posted your answer? My link to the older question was in the comments. – Gerry Myerson Nov 03 '13 at 00:46
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