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Sa se arate ca sirul \( (a_{n})_{n\geq1} \) dat prin formula:
\( a_{n}=\sum_{k=1}^{n}{\frac{1}{2^{k^2}}}\forall n \in\mathbb{N}* \)
nu poate fi scris sub forma \( a_{n}=\frac{f(n)}{g(n)}\forall n \in\mathbb{N}* \), unde \( f,g\in\mathbb{R}[X] \).


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Altfel ar rezulta ca \( a_{n+1}-a_n=\frac{1}{2^{(n+1)^2}}=\frac{f(n+1)}{g(n+1)}-\frac{f(n)}{g(n)} (\forall)n\in\mathbb{N^{\ast}} \)

Deoarece \( a_n \) este marginit(\( a_n\in (0,1) \)) si crescator rezulta ca este convergent deci f si g au acelasi grad.

Pe de alta parte ar rezulta ca \( \lim_{n\to \infty}\frac{(n+1)^2}{2^{(n+1)^2}}=0=\lim_{n\to\infty}(n+1)^2\frac{f(n+1)}{g(n+1)}-\frac{f(n)}{g(n)} \)

Presupunem ca exista cele doua polinoame . Ar rezulta ca

\( a_{n+1}-a_{n}=\frac{1}{2^{(n+1)^2}}=\frac{f(n+1)}{g(n+1)}-\frac{f(n)}{g(n)}=\frac{h(n)}{g(n+1)g(n)} \)
Fie sirul \( c_n=\frac{1}{2^{(n+1)^2}} \) de unde \( \frac{c_{n+1}}{c_n}=\frac{2^{(n+1)^2}}{2^{(n+2)^2}}=\frac{1}{2^{2n+3}}=\frac{h(n+1)}{g(n+2)g(n+1)}\cdot\frac{g(n)g(n+1)}{h(n)}=\frac{h(n+1)}{h(n)}\cdot\frac{g(n)}{g(n+2)} \)

Trecand la limita obtinem \( 0=1\cdot 1=1 \) contradictie.