Question

Calcule a exponencial

$\mathsf{e^{\,\sum\limits_{k=1}^{n} \ell n\left(1+\frac{1}{k}\right)}}$

e expresse a fórmula fechada em termos de n.

Answer 1

$\exp\bigg[\displaystyle\sum_{k=1}^{n}}\ln\bigg(1+\frac{1}{k}\bigg)\bigg]$

Usaremos a seguinte propriedade de logaritmos:

$\displaystyle\log_{b}\bigg(\prod_{k=p}^{n}a_{k}\bigg)=\sum_{k=p}^{n}\log_{b}(a_{k})$

(O logaritmo do produto é a soma dos logaritmos)

$\exp\bigg[\displaystyle\sum_{k=1}^{n}}\ln\bigg(1+\frac{1}{k}\bigg)\bigg]=\exp\bigg[\ln\,\prod_{k=1}^{n}\bigg(1+\frac{1}{k}\bigg)\bigg]$

Agora, usaremos $b^{\log_{b}(a)}=a~~\forall\,\,a,\,b~\textgreater~0,\,\,b\neq1$

$\displaystyle\exp\bigg[\sum_{k=1}^{n}}\ln\bigg(1+\frac{1}{k}\bigg)\bigg]=\prod_{k=1}^{n}\bigg(1+\frac{1}{k}\bigg)\\\\\\\exp\bigg[\sum_{k=1}^{n}}\ln\bigg(1+\frac{1}{k}\bigg)\bigg]=\prod_{k=1}^{n}\bigg(\dfrac{k+1}{k}\bigg)$

Como $\displaystyle\prod\limits_{k}\dfrac{a_{k}}{b_{k}}=\dfrac{\prod\limits_{k}a_{k}}{\prod\limits_{k}b_{k}}$:

$\displaystyle\exp\bigg[\sum_{k=1}^{n}}\ln\bigg(1+\frac{1}{k}\bigg)\bigg]=\dfrac{\prod\limits_{k=1}^{n}(k+1)}{\prod\limits_{k=1}^{n}k}\\\\\\\exp\bigg[\sum_{k=1}^{n}}\ln\bigg(1+\frac{1}{k}\bigg)\bigg]=\dfrac{(n+1)\prod\limits_{k=1}^{n-1}(k+1)}{1\cdot\prod\limits_{k=2}^{n}k}\\\\\\\exp\bigg[\sum_{k=1}^{n}}\ln\bigg(1+\frac{1}{k}\bigg)\bigg]=\dfrac{(n+1)\prod\limits_{k=2}^{n}k}{\prod\limits_{k=2}^{n}k}$

Então, temos que

$\boxed{\boxed{\exp\bigg[\sum_{k=1}^{n}\ln\bigg(1+\frac{1}{k}\bigg)\bigg]=n+1}}$