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I'm trying to test the summation $\sum^\infty_{n=0}\frac{(-1)^n}{5n+1}$ for absolute convergence.

By alternating series test, I can tell is is at least conditionally convergent.

However, when I used the ratio test, I got 1, which means it doesn't tell us anything.

A google search showed an answer on yahoo answers using the limit comparison test, using the harmonic series to compare it with, but they seemed to ignore the $(-1)^n$...

The answer in the book says it is conditionally convergent, but I can't work out how to show that it is not absolutely convergent.

Any ideas?

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  • $\begingroup$ If any of the answeres below were useful to you, then you should upvote all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this. $\endgroup$ – gebruiker Apr 6 '16 at 8:23
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Taking the absolute value and using the asymptotic comparison we find $$\frac1{5n+1}\sim_\infty \frac15\frac1n$$ that the series isn't absolutely convergent since the harmonic series isn't convergent.

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  • $\begingroup$ Short and sweet! $\endgroup$ – Namaste Apr 30 '14 at 12:16
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The absolute series would be: $\sum_{n=0}^\infty\frac{1}{5n+1}\geq\sum_{n=0}^\infty\frac{1}{6n}$ (at least for sufficiantly large $n$ it is). Then:$$\sum_{n=0}^\infty\frac{1}{6n}=\frac16\sum_{n=0}^\infty\frac{1}{n}$$ Wich is obviously divirgent.

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You have:

$\left|\frac{(-1)^n}{5n+1} \right|=\frac{1}{5n+1}>\frac{1}{5n+5}$,

therefore:

$\sum\limits_{n=0}^{\infty}\left|\frac{(-1)^n}{5n+1} \right|>\sum\limits_{n=0}^{\infty}\frac{1}{5n+5}=\frac{1}{5}\sum\limits_{n=1}^{\infty}\frac{1}{n}=\infty$,

So the series is not absolutely convergent.

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Taking the absolute value, we can apply the integral test:

$$\int_1^\infty \! \frac{1}{5x+1} dx = \lim_{N \to \infty}\frac{1}{5}\Big(\ln(5N+1) - \ln(6)\Big)$$

Certainly, the natural log is unbounded. Therefore, the integral does not converge, and we know the integral converges $\iff$ the series converges.

We conclude the series is not absolutely convergent. However, it is conditionally convergent by the alternating series test.

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