Question

Mostre que a série

$\mathsf{\displaystyle\sum_{k=0}^{\infty}}$ 1/[k! + (k + 1)!]

converge a 1.

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Dica: Faça análise da sequência das somas parciais. Ao encontrar um padrão, deduza uma possível fórmula fechada para a soma parcial e verifique se está correta (usando indução, por exemplo)

Answer 1

Vamos avaliar as três primeiras somas parciais:

$\mathsf{S_{0}=a_{0}=\dfrac{1}{0!+(0+1)!}=\dfrac{1}{1+1}=\dfrac{1}{2}}\\\\\\\mathsf{S_{1}=a_{0}+a_{1}=\dfrac{1}{2}+\dfrac{1}{1!+(1+1)!}=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{3+2}{2\cdot3}=\dfrac{5}{6}}\\\\\\\mathsf{S_{2}=a_{0}+a_{1}+a_{2}=S_{1}+a_{2}=\dfrac{5}{6}+\dfrac{1}{2!+(2+1)!}=\dfrac{5}{6}+\dfrac{1}{8}=\dfrac{23}{24}}$

Já é possível perceber um padrão: nos denominadores, temos os fatoriais de 2, 3 e 4 (nos índices 0, 1 e 2, respectivamente), e nos numeradores, temos o denominador diminuído de 1 unidade

Portanto, devemos verificar se

$\mathsf{S_{k}=\dfrac{(k+2)!-1}{(k+2)!},\,k\ge0}$

Faremos isso por indução:

Mostrando validade para $\mathsf{k=0}$:

$\mathsf{S_{0}=\dfrac{(0+2)!-1}{(0+2)!}=\dfrac{2!-1}{2!}=\dfrac{2-1}{2}=\dfrac{1}{2}~~~~\checkmark}$

Hipótese de indução: Assuma validade para $\mathsf{k=n~\textgreater~1}$, isto é,

$\mathsf{S_{n}=\dfrac{(n+2)!-1}{(n+2)!}}$

Mostrando que vale para $\mathsf{S_{n+1}}$:

$\mathsf{S_{n+1}=a_{0}+a_{1}+...+a_{n+1}}\\\\\\\mathsf{S_{n+1}=S_{n}+a_{n+1}}\\\\\\\mathsf{S_{n+1}=\dfrac{(n+2)!-1}{(n+2)!}+\dfrac{1}{(n+1)!+(n+1+1)!}}\\\\\\\mathsf{S_{n+1}=\dfrac{(n+2)!-1}{(n+2)!}+\dfrac{1}{(n+1)!+(n+2)!}}\\\\\\\mathsf{S_{n+1}=\dfrac{\big[(n+2)!-1\big]\cdot\big[(n+1)!+(n+2)!\big]+(n+2)!}{(n+2)!\cdot\big[(n+1)!+(n+2)!\big]}}\\\\\\\mathsf{S_{n+1}=\dfrac{(n+1)!(n+2)!+\big[(n+2)!\big]^{2}-(n+1)!-(n+2)!+(n+2)!}{(n+1)!\cdot(n+2)!+\big[(n+2)!\big]^{2}}}$

$\mathsf{S_{n+1}=\dfrac{\gamma(n)-(n+1)!}{\gamma(n)}}$

onde

$\mathsf{\gamma(n)=(n+1)!(n+2)!+[(n+2)!]^{2}=(n+2)!\cdot\big[(n+1)!+(n+2)!\big]}\\\\\mathsf{=(n+2)!\cdot\big[(n+1)!+(n+2)\cdot(n+1)!\big]=(n+1)!\cdot(n+2)!\big[1+n+2\big]}\\\\\mathsf{=(n+1)!\cdot(n+2)!\cdot(n+3)=(n+1)!\cdot(n+3)!}$

Voltando, temos

$\mathsf{S_{n+1}=\dfrac{(n+1)!\cdot(n+3)!-(n+1)!}{(n+1)!\cdot(n+3)!}}\\\\\\\mathsf{S_{n+1}=\dfrac{(n+1)!\cdot\big[(n+3)!-1\big]}{(n+3)!}}\\\\\\\boxed{\boxed{\mathsf_{S_{n+1}=\dfrac{(n+3)!-1}{(n+3)!}=\dfrac{(\mathbf{n+1}+2)!-1}{(\mathbf{n+1}+2)!}}}}$

Portanto, está provado, por indução, que $\mathsf{S_{k}=\frac{(k+2)!-1}{(k+2)!}~\forall\,k\ge0}$
__________________________

Finalmente, temos que

$\mathsf{S_{k}=\displaystyle\sum_{i=0}^{k}\dfrac{1}{i!+(i+1)!}=\dfrac{(k+2)!-1}{(k+2)!}}$

Por definição, $\displaystyle\mathsf{\sum_{k=0}^{\infty}a_{k}=\lim_{n\to\infty}\sum_{k=0}^{n}a_{k}}$

$\displaystyle\mathsf{\sum_{k=0}^{\infty}\dfrac{1}{k!+(k+1)!}=\lim_{n\to\infty}\sum_{k=0}^{n}\dfrac{1}{k!+(k+1)!}}\\\\\\\mathsf{\sum_{k=0}^{\infty}\dfrac{1}{k!+(k+1)!}=\lim_{n\to\infty}S_{n}}\\\\\\\mathsf{\sum_{k=0}^{\infty}\dfrac{1}{k!+(k+1)!}=\lim_{n\to\infty}\dfrac{(n+2)!-1}{(n+2)!}}\\\\\\\mathsf{\sum_{k=0}^{\infty}\dfrac{1}{k!+(k+1)!}=\lim_{n\to\infty}\bigg[1-\dfrac{1}{(n+2)!}\bigg]}\\\\\\\boxed{\boxed{\mathsf{\sum_{k=0}^{\infty}\dfrac{1}{k!+(k+1)!}=1}}}$