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Decide if the series converges absolutely, conditionally, or not at all. \begin{equation} \sum_{n=1}^{\infty}\frac{(-1)^n}{(2+(-1)^n)n} \end{equation}

I'm having a lot of trouble with this one. I know the even terms will be of the form $\frac{1}{3n}$ and the odd terms of the form $-\frac{1}{n}$ but I'm not really sure how I can use that to prove convergence. Any hints would be greatly appreciated.

Lucian
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user99219
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    Multiply by $-1$ in order to make things nicer, and let the resulting series be $a_1+a_2+a_3+\cdots$. Then the sum of the $(2n-1)$-th term and the $(2n)$-th term is $\frac{1}{2n-1}-\frac{1}{6n}$. Simplify. – André Nicolas Oct 03 '14 at 16:07
  • @AndréNicolas I think he meant $\sum_{n=1}^\infty\frac{(-1)^n}{(2+(-1)^n)n}$. – Troy Woo Oct 03 '14 at 16:13
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    I multiplied by $-1$ because of a dislike of negative numbers. Makes no difference to convergence. – André Nicolas Oct 03 '14 at 16:33

3 Answers3

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We have: $$\sum_{n=1}^{2N+1}\frac{(-1)^n}{(2+(-1)^n)n}=\frac{H_N}{6}-\sum_{k=0}^{N}\frac{1}{2k+1}=\frac{H_N}{6}-H_{2N+1}+\frac{H_N}{2}$$ hence: $$\sum_{n=1}^{2N+1}\frac{(-1)^n}{(2+(-1)^n)n}=\frac{1}{3}\log N-\log(2N)+O(1)=-\frac{2}{3}\log N+O(1)$$ so the series diverges towards $-\infty$.

Jack D'Aurizio
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This answer is based on Andre Nicolas's hint above:

$-S_{2n}=\displaystyle\sum_{k=1}^{n}\left(\frac{1}{2k-1}-\frac{1}{3(2k)}\right)$ where $\displaystyle\frac{1}{2k-1}-\frac{1}{3(2k)}>\frac{1}{3k}$, so

$-S_{2n}>\frac{1}{3}H_{n}\implies S_{2n}<-\frac{1}{3}H_{n}\implies\displaystyle\lim_{n\to\infty}S_{2n}=-\infty$ and hence the series diverges.

user84413
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I agree with your analysis of the problem: even terms will be of the form 1/3n and the odd terms of the form −1/n. Then each of them will converge to the same number 0. so the whole thing will converge to 0. Afterward, you can prove by induction.

Tesoro
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    The question is asking for the sum, not the limit, of the sequence. –  Oct 03 '14 at 19:43