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Continuing the answer given here: How to determine if this series converges absolutely/conditionally or diverges?

wrt to this series: $$ \sum_{n=2}^{\infty} \ln \left(1+\frac{(-1)^n}{n}\right) $$ will you please help me understand why is it legitimate to split this series into odd and even terms? i.e.- if the series indeed converges conditionally, we know that different splitting will give different values, right?

Where is my misunderstanding?

thanks

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    $\begingroup$ It is just the logarithm of Wallis' product. Anyway, splitting is allowed only for absolutely convergent series, and that is not the case. $\endgroup$ – Jack D'Aurizio Mar 29 '16 at 19:23
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You're right, that's an important step that should be in the answer to the question you linked. In this case, the terms of the series converge to zero, so we're okay. In more detail:

Let $a_k$ be the $k^{\text{th}}$ term, and let $S_k$ be the sum of the first $k$ terms. Your respondent showed that $S_{2k}=0$. That means that $S_{2k+1}$ is equal to $a_{2k+1}$. So, this shows that $0\leq|S_{k}|\leq|a_k|$ and $|a_k|$ goes to zero. So, by the squeeze theorem, $S_{k}$ goes to $0$.

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It can be bracketed as follows: $$\sum_{n=2}^{\infty}{\ln\left(1+\frac{(-1)^n}{n}\right)} \sim \sum_{k=1}^{\infty}{\ln\left(1+\frac{1}{2k} \right)} + \ln\left(1 - \frac{1}{2k+1}\right),$$ The last series can be simplified: $$\sum_{k=1}^{\infty}{\ln\left(1+\frac{1}{2k}\right)\left(1 - \frac{1}{2k+1}\right)} = \sum_{k=1}^{\infty}\ln1 = 0.$$ So the sum is equal to $0$. But $$\sum_{n=2}^{\infty}{|\ln\left(1+\frac{(-1)^n}{n}\right)|}$$ does not converge, so you cannot split it into odd and even terms.

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  • $\begingroup$ Many thanks !!!!! $\endgroup$ – Bobby Mar 30 '16 at 6:30

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