Question

Avalie a série infinita:
$\displaystyle \large{ \sum_{n = 1}^{ \infty} \frac{1}{(n + 1) \sqrt{n} + n \sqrt{n + 1} } }$

Por favor, mostre seu trabalho também - obrigado!​

Answer 1

• O valor da soma é igual a:

  $\Large\displaystyle\text{ \begin{gathered}\tt \sum ^{\infty}_{n=1} \frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=1\end{gathered} }$

Desejamos calcular a seguinte soma infinita:

  $\Large\displaystyle\text{ \begin{gathered}\tt \sum ^{\infty}_{n=1} \frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}\end{gathered} }$

Onde temos que, simplificando, ficamos da seguinte forma:

$\Large\displaystyle\text{ \begin{gathered}\tt \lim_{k \to \infty}\sum ^{k}_{n=1} \frac{(n+1)\sqrt{n} -n\sqrt{n+1}}{n(n+1)} \end{gathered} }$

Vamos então abrir em frações parciais, ficando então:

$\Large\displaystyle\text{ \begin{gathered}\tt \frac{(n+1)\sqrt{n} -n\sqrt{n+1}}{n(n+1)} =\frac{A}{n}+\frac{B}{n+1}\end{gathered} }$

$\Large\displaystyle\begin{gathered}\tt A(n+1)+Bn = (n+1)\sqrt{n} - n\sqrt{n+1} \end{gathered}$

Portanto, temos que:

$\Large\displaystyle\text{ \begin{gathered} \boxed{\tt A=\sqrt{n} }\ \ \wedge \ \ \boxed{\tt B=-\sqrt{n+1}}\end{gathered} }$

Substituindo:

$\Large\displaystyle\text{ \begin{gathered}\tt \lim_{k \to \infty}\sum ^{k}_{n=1} \frac{\sqrt{n}}{n} -\frac{\sqrt{n+1}}{n+1} \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered}\tt \sum ^{k}_{n=1} \frac{\sqrt{n}}{n} -\frac{\sqrt{n+1}}{n+1} = 1-\cancel{\frac{\sqrt{2}}{2}}+\cancel{\frac{\sqrt{2}}{2}}-\cancel{\ldots} +\cancel{\frac{\sqrt{k}}{k}}- \frac{\sqrt{k+1}}{k+1}\end{gathered} }$

Por fim:

$\Large\displaystyle\text{ \begin{gathered}\tt \lim_{k \to \infty}1-\frac{\sqrt{k+1}}{k+1} \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered}\tt 1-\underbrace{\tt \lim_{k \to \infty}\frac{\sqrt{k+1}}{k+1}}_{\longrightarrow 0} \end{gathered} }$

Portanto:

  $\Large\displaystyle\text{ \begin{gathered}\tt \sum ^{\infty}_{n=1} \frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=1\end{gathered} }$

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