Alternating series (Leibniz criterion)

In summary, the conversation discusses the conditions for an alternating series to converge, which include being monotonous and converging to zero. However, the conditions are sufficient but not necessary for convergence. The particular series being discussed, \Sigma_{n=1}^{\infty} (-1)^n | \frac{1}{n^2} \sin(n)|, can be proven to converge absolutely by comparison with 1/n^2. The confusion arises from a statement in class, but the "leftarrow" implication only implies what the thesis says and not the hypothesis.
  • #1
Damidami
94
0
I read that an alternating series [tex] \Sigma (-1)^n a_n [/tex] converges if "and only if" the sequence [tex]a_n [/tex] is both monotonous and converges to zero.

I tried with this series:

[tex] \Sigma_{n=1}^{\infty} (-1)^n | \frac{1}{n^2} \sin(n)| [/tex]

in the wolfram alpha and seems to converge to -0.61..., even if [tex] a_n = |\frac{1}{n^2} \sin(n)| [/tex] is not monotonous decreasing.

What am I doing wrong? Is the monotone condition necesary for this test, but the fact that a_n is not monotonous does not guarantee if the series converges or not?

Thanks.
 
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  • #2
The conditions given are sufficient, not necessary. It is possible for the terms to not be decreasing in magnitude, and for the whole series to still converge.

For your particular series, you can prove it converges absolutely by comparison with 1/n2
 
  • #3
Office_Shredder said:
The conditions given are sufficient, not necessary. It is possible for the terms to not be decreasing in magnitude, and for the whole series to still converge.

For your particular series, you can prove it converges absolutely by comparison with 1/n2

You are right, thanks! I confused because in class we saw this statement:

Let [tex] a_k > 0 [/tex] be a sequence monotonous decreasing. Then [tex] \Sigma_{n=1}^{\infty} (-1)^k a_k [/tex] converges [tex] \Leftrightarrow a_k \to 0 [/tex]

The [tex] \Rightarrow [/tex] part of the proof is simply the necesary condition for any series to converge that [tex] a_k \to 0 [/tex].

But I thoght the [tex] \Leftarrow [/tex] part would imply everything besides it: both that the series converges and that [tex] a_k [/tex] is monotonous decreasing.

So it's a logical question now: when we have a double implication like above, the "leftarrow" implication only implies what the thesis says, but the hipothesis is the same. Am I right now?

Thanks!
 

1. What is the Leibniz criterion for an alternating series?

The Leibniz criterion is a test used to determine the convergence or divergence of an alternating series. It states that if the terms of an alternating series decrease in absolute value and approach zero, then the series is convergent.

2. How is the Leibniz criterion different from other convergence tests?

The Leibniz criterion is specifically used for alternating series, while other convergence tests such as the ratio and root tests can be applied to both alternating and non-alternating series.

3. Can the Leibniz criterion be used to determine the exact value of a convergent alternating series?

No, the Leibniz criterion only determines the convergence or divergence of an alternating series. To find the exact value of a convergent series, other methods such as the partial sum formula or Taylor series expansion must be used.

4. What is the role of the alternating series test in the Leibniz criterion?

The alternating series test is a necessary condition for the Leibniz criterion to be applicable. It states that the terms of the series must alternate in sign and decrease in absolute value for the Leibniz criterion to be valid.

5. Are there any exceptions to the Leibniz criterion?

Yes, there are some cases where the Leibniz criterion may not accurately determine the convergence or divergence of an alternating series. These exceptions include series with non-decreasing terms, or series where the terms do not approach zero.

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