# Absolute or Conditional Convergence, or Divergence of Alternating Series.

• icesalmon
In summary, the conversation discussed determining the convergence of a given infinite series and the conditions for absolute and conditional convergence. The series in question was found to be conditionally convergent using the alternating series test, but not absolutely convergent. The importance of correctly identifying the type of convergence was also emphasized.
icesalmon

## Homework Statement

given an = ( -1 )(n+1) / $\sqrt{n}$ determine if the infinite series is Absolutely Convergent, Conditionally Convergent, or Divergent.

## Homework Equations

I hope I have these theorems down correctly, please correct me if I'm wrong. If $\Sigma$|an| is Convergent then $\Sigma$an is Absolutely Convergent if $\Sigma$|an| is Divergent and $\Sigma$an is Convergent then $\Sigma$an is Conditionally Convergent And if neither occur then the $\Sigma$an may be Divergent.

## The Attempt at a Solution

To show $\Sigma$|an| or $\Sigma$an are convergent I have to show 2 things, 1). lim[ n$\rightarrow$$\infty$ ] |an| = 0 and that either an+1 $\leq$ an, an+1/an $\leq$ 1 or d(an)/dn < 0 which determine whether or not an is Decreasing. If I can show these things I have a convergent Alternating Series, if not I might be able to try another test or I have a divergent Alternating Series.

So I started by seeing if lim[ n $\rightarrow$ $\infty$ ] |an| = 0 so I have to figure out what |an| actually is. Again correct me if I'm wrong, it would help me out greatly here, but I believe |(-1)n+1/$\sqrt{n}$| to be 1/$\sqrt{n}$ because |+/-(1)| = 1 there for I can determine that lim[ n $\rightarrow$ $\infty$ ] 1/$\sqrt{n}$ = 1/$\infty$ = 0 so if I have my thoughts in order I have shown lim[n $\rightarrow$ $\infty$ ] 1/$\sqrt{n}$ = 0 great! Next I have to show that an+1 $\leq$ an so I first considered $\sqrt{n+1}$ which$\geq$ $\sqrt{n}$ so 1/|$\sqrt{n+1}$| $\leq$ 1/|$\sqrt{n}$|
to further solidify this in my mind or at least on paper, I wanted to show that d(|an|/dn $\leq$ 0 so I did just that and acquired -1/(2n3/2) $\leq$ 0 $\forall$ $\Re$ therefor $\Sigma$an is Absolutely Convergent. The book says that it is conditionally convergent, I tried the integral test for no good reason and paired with my above results got the exact opposite of my definition for conditional convergence.

After moving on I got the subsequent problems incorrect as well, falsly concluding that my series was Absolutely Convergent when they were Conditionally Convergent so I can't really move anywhere until I understand this, thanks in advance for your help.

i think you're on the right track, the series is conditionally convergent shown by applying the "alternating series test"
-an tends to zero monotonically and changes sign
so the series as its written it converges to a finite limit

however as you've also correctly worked out it is not absolutely convergent as the sum of the modulus of each term diverges

any absolutely convergent series is conditionally convergent, however the converse does not apply

## 1. What is the difference between absolute and conditional convergence?

Absolute convergence refers to the convergence of a series where the absolute values of the terms in the series are used. Conditional convergence, on the other hand, refers to the convergence of a series where the signs of the terms are taken into account. In other words, a series that is absolutely convergent will also be conditionally convergent, but the opposite may not always be true.

## 2. How can I determine if an alternating series is convergent or divergent?

The alternating series test is commonly used to determine the convergence or divergence of an alternating series. This test states that if the absolute values of the terms in the series decrease monotonically and approach zero, then the series is convergent. If the absolute values of the terms do not approach zero, then the series is divergent.

## 3. Can an alternating series be both absolutely and conditionally convergent?

Yes, it is possible for an alternating series to be both absolutely and conditionally convergent. This means that the series converges regardless of whether the signs of the terms are taken into account or not. An example of such a series is the alternating harmonic series, where both the series and its absolute values converge to the same value.

## 4. How does the alternating series test differ from the divergence test?

The alternating series test and the divergence test are two different tests used to determine the convergence or divergence of a series. The divergence test states that if the limit of the terms in a series does not approach zero, then the series is divergent. The alternating series test, on the other hand, is specifically used for alternating series and looks at the absolute values of the terms to determine convergence or divergence.

## 5. Are there any other tests for determining the convergence or divergence of alternating series?

Yes, there are other tests, such as the ratio test and the integral test, that can be used to determine the convergence or divergence of alternating series. These tests may be more useful for certain types of series and can provide a quicker and more definitive answer. However, the alternating series test is specifically designed for alternating series and can be a useful tool for determining convergence or divergence in those cases.

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