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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 ?

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    $\begingroup$ yes, it is correct. We can show the same following the hint of DonAntonio. $\endgroup$ – Masacroso Apr 15 '17 at 14:25
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This is correct. Sorry I can't say more but you got all the steps

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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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So your function is:

$f(x)=\dfrac{(-1)^n * 2}{n+1}$

since $f(n)$ as $x$ approaches infinity, is $0$, the function is not divergent. So it's either conditionally convergent or absolutely convergent.

To check if it's absolute or not, we have to check if $|f(x)|$ is convergent or not.

$f(n)=\dfrac{2}{n+1}$

With the integral test we can understand that this function diverges. So $\sum_{k=1}^{\infty}f(n)$ is conditionally convergent.

$\int_{2}^{\infty}\dfrac{1}{n+1}=\lim_{b\rightarrow\infty}2\ln(\dfrac{b+1}{2})\rightarrow\infty$

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As said by DonAntonio, the general term of the series is

$$u_n=2\frac {(-1)^n}{n+1} $$

and

$$|u_n|=\frac {2}{n+1} $$ thus

$\sum u_n $ is convergent as alternate series and $\sum |u_n|$ diverges as a Riemann series.

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