Question

(50 PONTOS) Obter uma forma fechada para as somas, utilizando a propriedade telescópica:

a) $\displaystyle\sum\limits_{n=0}^{k}{2^{(n^2)}\cdot (2^{2n+1}-1)}$

b) $\displaystyle\sum\limits_{n=2}^{k}{\mathrm{\ell n}\left[1+\dfrac{\mathrm{\ell n}(n+1)}{\mathrm{\ell n}(n!)} \right ]}$

Answer 1

a)

$2^{(n^{2})}\cdot(2^{2n+1}-1)=2^{(n^{2})}\cdot2^{2n+1}-1\cdot2^{(n^{2})}\\\\2^{(n^{2})}\cdot(2^{2n+1}-1)=2^{n^{2}+2n+1}-2^{(n^{2})}\\\\2^{(n^{2})}\cdot(2^{2n+1}-1)=2^{(n+1)^{2}}-2^{(n^{2})}$

Definindo a sequência

$a_{n}=2^{(n^{2})}$

Temos que

$2^{(n^{2})}\cdot(2^{2n+1}-1)=\Delta(a_{n})$

Portanto:

$\displaystyle\sum\limits_{n=0}^{k}2^{(n^{2})}\cdot(2^{2n+1}-1)=\sum\limits_{n=0}^{k}\Delta(a_{n})\\\\\\\sum\limits_{n=0}^{k}2^{(n^{2})}\cdot(2^{2n+1}-1)=a_{k+1}-a_{0}\\\\\\\sum\limits_{n=0}^{k}2^{(n^{2})}\cdot(2^{2n+1}-1)=2^{(k+1)^{2}}-2^{(0^{2})}\\\\\\\boxed{\boxed{\sum\limits_{n=0}^{k}2^{(n^{2})}\cdot(2^{2n+1}-1)=2^{(k+1)^{2}}-1}}$
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b)

$1+\dfrac{ln(n+1)}{ln(n!)}=\dfrac{ln(n!)}{ln(n!)}+\dfrac{ln(n+1)}{ln(n!)}\\\\\\1+\dfrac{ln(n+1)}{ln(n!)}=\dfrac{ln(n!)+ln(n+1)}{ln(n!)}\\\\\\1+\dfrac{ln(n+1)}{ln(n!)}=\dfrac{ln[n!\cdot(n+1)]}{ln(n!)}\\\\\\1+\dfrac{ln(n+1)}{ln(n!)}=\dfrac{ln[(n+1)!]}{ln(n!)}$

Então

$ln\left[1+\dfrac{ln(n+1)}{ln(n!)}\right]=ln\left[\dfrac{ln[(n+1)!]}{ln(n!)}\right]\\\\\\ln\left[1+\dfrac{ln(n+1)}{ln(n!)}\right]=ln(ln[(n+1)!])-ln(ln(n!))$

Se definirmos a sequência $a_{n}=ln(ln(n!))$, temos

$ln\left[1+\dfrac{ln(n+1)}{ln(n!)}\right]=\Delta(a_{n})$

Portanto:

$\displaystyle\sum\limits_{n=2}^{k}ln\left[1+\dfrac{ln(n+1)}{ln(n!)}\right]=\sum\limits_{n=2}^{k}\Delta(a_{n})\\\\\\\displaystyle\sum\limits_{n=2}^{k}ln\left[1+\dfrac{ln(n+1)}{ln(n!)}\right]=a_{k+1}-a_{2}\\\\\\\displaystyle\sum\limits_{n=2}^{k}ln\left[1+\dfrac{ln(n+1)}{ln(n!)}\right]=ln(ln([k+1]!))-ln(ln(2!))\\\\\\\boxed{\boxed{\displaystyle\sum\limits_{n=2}^{k}ln\left[1+\dfrac{ln(n+1)}{ln(n!)}\right]=ln(ln([k+1]!))-ln(ln(2))}}$

OBS: ln(ln(2)) está bem definido, pois ln(2) é positivo