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Let $$\sum_{n=1}^{\infty}{(1-2c^2)^n \over n \ln{n} + \cos{n\pi}}$$ $c \in \mathbb{R}$. For what values of $c$ is this series:

1) convergent?

2) absolutely convergent?

Do you have any suggestions?

  • 0
    Do you have any suggestions?2017-01-12

2 Answers 2

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Hint: concentrate on the numerator. Do a case division $|c|>,=,<\sqrt{2}/2$. The denominator is pretty much irrelevant

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Considering: $$ n\log{n}+\cos{n\pi} = n\log{n}+(-1)^n \approx n\log{n} \quad\{\text{for}\,n\rightarrow\infty\} \quad\qquad\qquad\qquad\qquad\qquad $$ Let: $$\color{red}{A=1-2c^2} \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad $$ Thus: $$ \sum_{n=\color{red}{2}}^{\infty}\frac{A^n}{n\log{n}} = \begin{cases} -\infty\lt \,A\, \lt-1 &\Rightarrow \sum\frac{(-1)^n|A|^n}{n\log{n}} \space\space\space\quad\rightarrow\quad \text{Diverge by Limit test} \\[2mm] -1\,\,\lt \,A\, \lt0 &\Rightarrow \sum\frac{(-1)^n}{(1/|A|)^n n\log{n}} \space\rightarrow\quad \text{Converge by Ratio test} \\[2mm] \,\,\,0\,\,\,\lt \,A\, \lt+1 &\Rightarrow \sum\frac{1}{(1/|A|)^n n\log{n}} \space\rightarrow\quad \text{Converge by Ratio test} \\[2mm] +1\,\,\lt \,A\, \lt+\infty &\Rightarrow \sum\frac{|A|^n}{n\log{n}} \space\space\space\qquad\rightarrow\quad \text{Diverge by Limit test} \\[2mm] \end{cases} $$ Also: $$ \sum_{n=\color{red}{2}}^{\infty}\frac{A^n}{n\log{n}} = \begin{cases} A=-1 &\implies \sum\frac{(-1)^n}{n\log{n}} \quad\rightarrow\quad \text{Converge by Dirichlet test} \\[2mm] A=0 &\implies \sum\frac{(0)^n}{n\log{n}} = 0 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \\[2mm] A=+1 &\implies \sum\frac{1}{n\log{n}} \quad\rightarrow\quad \text{Diverge by Comparison test} \\[2mm] \end{cases} $$ Hence: $$ \sum_{n=1}^{\infty}\frac{\left(1-2c^2\right)^n}{n\log{n}+\cos{n\pi}} = \begin{cases} \text{Convergent for} &A\in\,\color{blue}{[}-1,\,+1\color{red}{[} \quad\rightarrow\quad c\in\,\color{blue}{[}-1,\,+1\color{blue}{]}-\color{red}{\{0\}} \\[2mm] \text{|Convergent| for} &A\in\,\color{blue}{]}-1,\,+1\color{red}{[} \quad\rightarrow\quad c\in\,\color{blue}{]}-1,\,+1\color{blue}{[}-\color{red}{\{0\}} \\[2mm] \end{cases} $$

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