<s>[INST] Does the series $\sum_{n=1}^{\infty} \frac{(-1)^n - (-1)^{n+1}}{n+1}$ converges absolutely or conditionally? 
Does the following series converges absolutely or conditionally? $$\sum_{n=1}^{\infty} \frac{(-1)^n - (-1)^{n+1}}{n+1}$$

$$\sum_{n=1}^{\infty} \frac{(-1)^n - (-1)^{n+1}}{n+1}=\sum_{n=1}^{\infty} \frac{(-1)^n}{n+1} - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$$
Both $$\sum_{n=1}^{\infty} \frac{(-1)^n}{n+1}$$ and $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$$ are convergent by alternating test, hence $\sum_{n=1}^{\infty} \frac{(-1)^n}{n+1} - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$ is convergent.
$\sum_{n=1}^{\infty} \frac{(-1)^n - (-1)^{n+1}}{n+1}$ is convergent.
$$\sum_{n=1}^{\infty}| \frac{(-1)^n - (-1)^{n+1}}{n+1} = \sum_{n=1}^{\infty} \frac{2}{n+1}$$ which is divergent by comparison test with $$\sum_{n=1}^{\infty} \frac{1}{n}$$
Hence conditionally convergent.
Is this correct ?
 [/INST] Hint:
$$(-1)^n-(-1)^{n+1}=(-1)^n+(-1)^{n+2}=(-1)^{n}\left[1+1^2\right]=2\cdot(-1)^n\ldots$$
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