the series ((-1)^(n-1))/(n^p)
Use the alternating series test. If: 1) Each term is less than the previous term 2) The limit of the absolute value of the terms is 0 Then the series converges. Here we need 1/(n+1)^p <= 1/n^p, Isso só acontece se p >= 1. We also need the absolute value to be related to 0. This happens again when p >= 1. So for p >= 1 we know that the series converges by the alternating series test. Now if p < 1, podemos ver que o limite do termo geral não vai para 0. O termo n^p fica muito pequeno quando n fica grande, então a função oscila entre grandes valores positivos e grandes valores negativos, assim o limite não existe. Então sabemos que a soma diverge. A série converge se e somente se p >= 1.
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For what values of p is the following series convergent?
the series ((-1)^(n-1))/(n^p)
Let us take the series of the absolute values of the terms, that is to say the series 1/n^p. We know that this is a p-series, convergent exactly where p > 1. So the given series will *absolutely* be convergent exactly where p > 1.
Where is *conditionally* convergent? Since 1 / n^p monotonically converges to 0, the alternating series test tells us that (-1)^(n – 1) / n^p will converge. This will converge to 0 as n^p grows monotonically to infinity, which will happen exactly when p > 0. So, for p > 0, by the alternating series test, the series will converge.
And if p = 0? So the series is (-1)^(n – 1) / n^0 = (-1)^(n – 1) which does not converge.
And if p < 0? The series is therefore (-1)^(n - 1) * n^(-p), which will approach infinity. By the divergence test, the series does not converge. Therefore, the series is: 1) Divergent if p <= 0 2) Conditionally convergent on 0 < p <= 1 3) Absolute convergent on 1 < p
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