Question

(50 PONTOS) Mostre que
$\displaystyle\sum\limits_{n=1}^{k}{\mathrm{\ell n}\left(1+\dfrac{1}{n} \right )}=\mathrm{\ell n}(k+1)$

Answer 1

Vamos manipular o termo geral do somatório:

$ln\left(1+\dfrac{1}{n}\right)=ln\left(\dfrac{n}{n}+\dfrac{1}{n}\right)\\\\\\ln\left(1+\dfrac{1}{n}\right)=ln\left(\dfrac{n+1}{n}\right)\\\\\\\boxed{\boxed{ln\left(1+\dfrac{1}{n}\right)=ln(n+1)-ln(n)}}$

Então:

$\displaystyle\sum\limits_{n=1}^{k}ln\left(1+\dfrac{1}{n}\right)=\sum\limits_{n=1}^{k}[ln(n+1)-ln(n)]\\\\\\\sum\limits_{n=1}^{k}ln\left(1+\dfrac{1}{n}\right)=\sum\limits_{n=1}^{k}ln(n+1)-\sum\limits_{n=1}^{k}ln(n)$

Desenvolvendo os somatórios, temos

$\displaystyle\sum\limits_{n=1}^{k}ln\left(1+\dfrac{1}{n}\right)=ln(2)+ln(3)+...+ln(k+1)-(ln(1)+ln(2)+...+ln(k))\\\\\\=(0-ln(1))+(ln(2)-ln(2))+...+(ln(k)-ln(k))+ln(k+1)\\\\\\\sum\limits_{n=1}^{k}ln\left(1+\dfrac{1}{n}\right)=(0-0)+0+0+...+0+ln(k+1)\\\\\\\boxed{\boxed{\sum\limits_{n=1}^{k}ln\left(1+\dfrac{1}{n}\right)=ln(k+1)}}$