Disciplina de Cálculo IV do ICMC
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filipe.lemos
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by filipe.lemos »
\(\sum_{n=1}^{\infty} \frac{z^{n}}{a^{n}+ b^{n}}\)
1) Assumindo:
- \(b > a\) (lembrando que a mesma lógica deve ser aplicada para a hipótese contrária \(a > b)\):
\(\lim\limits_{n \rightarrow +\infty} \frac{C_{n+1}}{C_n} = \lim\limits_{n \rightarrow +\infty} \frac{\frac{1}{a^{n+1}+b^{n+1}}}{\frac{1}{a^{n}+b^{n}}} = \lim\limits_{n \rightarrow +\infty} \frac{\frac{a^{n}}{b^{n}}+\frac{b^{n}}{b^{n}}}{\frac{a.a^{n}}{b^{n}}+\frac{b.b^{n}}{b^{n}}} = \lim\limits_{n \rightarrow +\infty} \frac{1+(\frac{a}{b})^{n}}{a.(\frac{a}{b})^{n}+ b} = \frac{1}{b}\)
\(\therefore\) Se
\(b > 1 \to\) converge, mas se
\(b < 1 \to\) diverge.
2) Por outro lado, assumindo:
\(\sum_{n=1}^{\infty} \frac{z^{n}}{a^{n}+ a^{n}} = \sum_{n=1}^{\infty} \frac{z^{n}}{2.a^{n}} = \frac{1}{2}.\sum_{n=1}^{\infty} (\frac{z}{a})^{n}\)
\(\lim\limits_{n \rightarrow +\infty} \frac{C_{n+1}}{C_n} = \lim\limits_{n \rightarrow +\infty} \frac{\frac{1}{a^{n+1}}}{\frac{1}{a^{n}}} = \lim\limits_{n \rightarrow +\infty} \frac{1}{a} = \frac{1}{a}\)
\(\therefore\) Se
\(a > 1 \to \) converge, mas se
\(a < 1 \to\) diverge.