Question

Sa se calculeze sumele:
a) Sn = 1/(0!+1!) + 1/(1!+2!) + ... + 1/[(n-1)!+n!]
b) Sn = 3/(1!+2!+3!) + 4/(2!+3!+4!) + ... + (n+2)/[n!+(n+1)!+(n+2)!]

Answer 1

$a) \quad S_n = \sum\limits_{k=1}^{n}\dfrac{1}{(k-1)!+ k!} = \sum\limits_{k=1}^n \dfrac{1}{\dfrac{k!}{k} + k!} = \\ \\ = \sum\limits_{k=1}^{n} \dfrac{1}{k!\Big(\dfrac{1}{k}+1\Big)} = \sum\limits_{k=1}^{n} \dfrac{1}{k!\cdot \dfrac{k+1}{k}} = \sum\limits_{k=1}^{n} \dfrac{k}{k!(k+1)} = \\ \\ = \sum\limits_{k=1}^{n}\dfrac{k+1-1}{k!(k+1)} = \sum\limits_{k=1}^{n}\left(\dfrac{k+1}{k!(k+1)} - \dfrac{1}{k!(k+1)}\right) =$

$\\ \\ = \sum\limits_{k=1}^{n} \left(\dfrac{1}{k!} - \dfrac{1}{(k+1)!}\right) = \sum\limits_{k=1}^n\dfrac{1}{k!} - \sum\limits_{k=1}^{n} \dfrac{1}{(k+1)!} = \\ \\ = \dfrac{1}{1!} + \dfrac{1}{2!} + ... +\dfrac{1}{n!} - \left( \dfrac{1}{2!} + \dfrac{1}{3!} + ... + \dfrac{1}{n!} + \dfrac{1}{(n+1)!} \right) = \\ \\ = \dfrac{1}{1!} - \dfrac{1}{(n+1)!} = \boxed{1 - \dfrac{1}{(n+1)!}}$


$b) \quad S_n = \sum\limits_{k=1}^{n} \dfrac{k+2}{k!+(k+1)!+(k+2)!} = \\ \\ = \sum\limits_{k=1}^{n}\dfrac{k+2}{k!+k!(k+1)+(k+2)!} = \sum\limits_{k=1}^{n} \dfrac{k+2}{k!(1+k+1)+(k+2)!} = \\ \\ = \sum\limits_{k=1}^{n} \dfrac{k+2}{k!(k+2)+(k+1)!(k+2)}= \sum\limits_{k=1}^{n} \dfrac{k+2}{(k+2)\Big(k!+(k+1)!\Big)} =$

$\\ \\ = \sum\limits_{k=1}^{n} \dfrac{1}{k!+(k+1)!} = \sum\limits_{k=2}^{n+1} \dfrac{1}{(k-1)!+k!} \overset{\text{din a)}}{=} \sum\limits_{k=2}^{n+1} \dfrac{1}{k!} - \sum\limits_{k=2}^{n+1} \dfrac{1}{(k+1)!} =$

$=\dfrac{1}{2!} + \dfrac{1}{3!} + ... + \dfrac{1}{(n+1)!} - \left(\dfrac{1}{3!}+ \dfrac{1}{4!} + ... + \dfrac{1}{(n+1)!} + \dfrac{1}{(n+2)!}\right) = \\ \\ = \dfrac{1}{2!} - \dfrac{1}{(n+2)!} = \boxed{\dfrac{1}{2} - \dfrac{1}{(n+2)!}}$