Question

Determine if the series is absolutely convergent or conditionally convergent. Sum from n=0 to infinity ((-1)^n)/ (8n+1)

Answer 1

\[\sum_{n=1}^{\infty} \frac{ (-1)^{n} }{ 8n+1 }\]

Answer 2

Does its absolute part \(\left|\dfrac{(-1)^n}{8n+1}\right|=\dfrac1{8n+1}\) converge on its own? Hm, well you can use a comparison test with a harmonic series to show it doesn't -- so, no, it does not absolutely converge.

To see that it conditionally converges, we just show the absolute part tends to \(0\):$$\lim_{n\to\infty}\frac1{8n+1}=0$$... and that our absolute part also is monotone decreasing; this is trivial to show: \(n\lt n+1\) so \(8n\lt 8(n+1)\) and thus \(8n+1\lt8(n+1)+1\). Taking the reciprocal of both sides (paying attention to the direction of our inequality) we have \(\frac1{8n+1}\gt\frac1{8(n+1)+1}\) and therefore we have a decreasing function (you may also use the derivative and show that it is negative for \(n\gt k\) for some integer \(k\) as all terms before \(k\) can be broken off of our sum, as they are finite and do not affect infinite convergence).

Thus by the alternating series test our series does indeed converge conditionally.

Answer 3

I understand that the limit goes to 0 But I still am not following why its absolute does not and goes to infinity

Answer 4

Use Leibniz's theorem for alternating series, which says that a alternating seies \(\Sigma (-1)^na_n\) is convergent if \(a_{n+1}<a_n\) and \(\lim a_n =0 \). It will be also absolutly convergent if the series \(\Sigma |a_n|\) is convergent.

Answer 5

@SarahHow its absolute value series fails to converge by the limit comparison test; consider \(\dfrac1{8n}\). We then show \(0<C<\infty\) where $$C=\lim_{n\to\infty}\frac{\frac1{8n+1}}{\frac1{8n}}=\lim_{n\to\infty}\frac{8n}{8n+1}=\lim_{n\to\infty}\left(1-\frac1{8n+1}\right)=1$$... so, indeed, \(0<C<\infty\). By the limit comparison test, we see that either both converge or both diverge. Since \(\sum\limits_{n=0}^\infty\dfrac1{8n}=\dfrac18\sum\limits_{n=0}^\infty\dfrac1n\), i.e. a constant multiple of the harmonic series, it clearly diverges and thus so does our absolute series.

Answer 6

For an intuitive understanding, consider that when alternating, much of the values end up cancelling between terms as they alternate in sign... hence you typically see oscillaiton like: