Question

(30 PONTOS) Mostre que a série

$\displaystyle\sum\limits_{n=2}^{\infty}\dfrac{\mathrm{\ell n}\left(1+\frac{1}{n} \right )}{\mathrm{\ell n}(n+1)\cdot \mathrm{\ell n}(n)}$

converge e indique seu o valor da soma desta série.

Answer 1

Definindo a sequência numérica

$a_{n}=\dfrac{ln(1+\frac{1}{n})}{ln(n+1)\cdot ln(n)}$

Temos que

$b_{n}=\dfrac{ln(\frac{n+1}{n})}{ln(n+1)\cdot ln(n)}\\\\\\b_{n}=\dfrac{ln(n+1)-ln(n)}{ln(n+1)\cdot ln(n)}\\\\\\b_{n}=\dfrac{ln(n+1)}{ln(n+1)\cdot ln(n)}-\dfrac{ln(n)}{ln(n+1)\cdot ln(n)}\\\\\\b_{n}=\dfrac{1}{ln(n)}-\dfrac{1}{ln(n+1)}$

Se definirmos a sequência

$a_{n}=\dfrac{1}{ln(n)}$

Temos que

$\Delta(a_{n})=a_{n+1}-a_{n}=\dfrac{1}{ln(n+1)}-\dfrac{1}{ln(n)}=-b_{n}\\\\\\\boxed{\boxed{b_{n}=-\Delta(a_{n})}}$

Portanto, a k-ésima soma parcial da série dada é

$\displaystyle\sum\limits_{n=2}^{k}\dfrac{ln(1+\frac{1}{n})}{ln(n+1)\cdot ln(n)}=\sum\limits_{n=2}^{k}\left(\dfrac{1}{ln(n)}-\dfrac{1}{ln(n+1)}\right)\\\\\\\displaystyle\sum\limits_{n=2}^{k}\dfrac{ln(1+\frac{1}{n})}{ln(n+1)\cdot ln(n)}=\sum\limits_{n=2}^{k}-\Delta(a_{n})\\\\\\\displaystyle\sum\limits_{n=2}^{k}\dfrac{ln(1+\frac{1}{n})}{ln(n+1)\cdot ln(n)}=-\sum\limits_{n=2}^{k}\Delta(a_{n})$

E, como vimos

$\displaystyle\sum\limits_{n=2}^{k}\dfrac{ln(1+\frac{1}{n})}{ln(n+1)\cdot ln(n)}=-(a_{k+1}-a_{2})\\\\\\\displaystyle\sum\limits_{n=2}^{k}\dfrac{ln(1+\frac{1}{n})}{ln(n+1)\cdot ln(n)}=a_{2}-a_{k+1}\\\\\\\displaystyle\sum\limits_{n=2}^{k}\dfrac{ln(1+\frac{1}{n})}{ln(n+1)\cdot ln(n)}=\dfrac{1}{ln(2)}-\dfrac{1}{ln(k+1)}$

Por definição, se o limite

$\displaystyle\lim\limits_{n\rightarrow\infty}\sum\limits_{n=2}^{k}b_{n}$

existir, temos que a série converge e seu valor é definido como o valor do limite

$\lim\limits_{n\rightarrow\infty}\displaystyle\sum\limits_{n=2}^{k}\dfrac{ln(1+\frac{1}{n})}{ln(n+1)\cdot ln(n)}=\lim\limits_{n\rightarrow\infty}\left(\dfrac{1}{ln(2)}-\dfrac{1}{ln(k+1)}\right)$

Como $\dfrac{1}{ln(k+1)}\longrightarrow0~quando~k\longrightarrow\infty$

Portanto:

$\displaystyle\sum\limits_{n=2}^{\infty}\dfrac{ln(1+\frac{1}{n})}{ln(n+1)\cdot ln(n)}=\dfrac{1}{ln(2)}-0\\\\\\\boxed{\boxed{\sum\limits_{n=2}^{\infty}\dfrac{ln(1+\frac{1}{n})}{ln(n+1)\cdot ln(n)}=\dfrac{1}{ln(2)}}}$