Question

Mostre que a série a seguir converge a 1:
$\displaystyle\sum_{k=0}^{\infty}\frac{1}{(k+1)!+k!}$

Answer 1

Vamos manipular o somando:

$b_{k}=\dfrac{1}{(k+1)!+k!}=\dfrac{1}{(k+1)\cdot k!+k!}\\\\\\\dfrac{1}{(k+1)!+k!}=\dfrac{1}{[k+1+1]\cdot k!}\\\\\\\dfrac{1}{(k+1)!+k!}=\dfrac{1}{(k+2)\cdot k!}$

Multiplicando o numerador e o denominador por $(k+1)\cdot(k+1)!$:

$\dfrac{1}{(k+1)!+k!}=\dfrac{(k+1)\cdot(k+1)!}{(k+2)\cdot(k+1)!\cdot(k+1)\cdot k!}\\\\\\\dfrac{1}{(k+1)!+k!}=\dfrac{(k+1)\cdot(k+1)!}{(k+2)!\cdot(k+1)!}=b_{k}$

Usando um raciocínio análogo ao utilizado para resolver a tarefa http://brainly.com.br/tarefa/7251949, encontramos uma sequência

$a_{k}=-\dfrac{1}{(k+1)!}$

tal que

$\Delta(a_{k})=a_{k+1}-a_{k}\\\\\Delta(a_{k})=-\dfrac{1}{(k+2)!}+\dfrac{1}{(k+1)!}\\\\\\\Delta(a_{k})=\dfrac{k+2}{(k+2)(k+1)!}-\dfrac{1}{(k+2)!}\\\\\\\Delta(a_{k})=\dfrac{k+2-1}{(k+2)!}=\dfrac{k+1}{(k+2)!}=\dfrac{(k+1)\cdot(k+1)!}{(k+2)\cdot(k+1)!}=b_{k}$

Portanto, temos

$\displaystyle\sum\limits_{k=0}^{n}b_{k}=\sum\limits_{k=0}^{n}\Delta(a_{k})\\\\\\\sum\limits_{k=0}^{n}b_{k}=a_{n+1}-a_{0}\\\\\\\sum\limits_{k=0}^{n}\dfrac{1}{(k+1)!+k!}=-\dfrac{1}{(n+2)!}+\dfrac{1}{(1+0)!}\\\\\\\boxed{\boxed{\sum\limits_{k=0}^{n}\dfrac{1}{(k+1)!+k!}=1-\dfrac{1}{(n+2)!}}}$

Logo, por definição:

$\displaystyle\sum\limits_{k=0}^{\infty}\dfrac{1}{(k+1)!+k!}=\lim\limits_{n\to\infty}\sum\limits_{k=0}^{n}\dfrac{1}{(k+1)+k!}\\\\\\\sum\limits_{k=0}^{\infty}\dfrac{1}{(k+1)!+k!}=\lim\limits_{n\to\infty}\bigg[1-\dfrac{1}{(n+2)!}\bigg]\\\\\\\sum\limits_{k=0}^{\infty}\dfrac{1}{(k+1)!+k!}=1-0\\\\\\\boxed{\boxed{\sum\limits_{k=0}^{\infty}\dfrac{1}{(k+1)!+k!}=1}}$