# How do you test for convergence of #Sigma (-1)^n(1-n^2)# from #n=[1,oo)#?

Diverges

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To test for the convergence of the series (\sum_{n=1}^{\infty} (-1)^n(1-n^2)), we can use the alternating series test. This test states that if a series alternates signs and the absolute value of its terms decreases as (n) increases, and if the limit of the absolute value of the terms approaches zero as (n) approaches infinity, then the series converges.

In this series, the terms alternate signs and the absolute value of the terms ((1-n^2)) is increasing as (n) increases. However, the limit of ((1-n^2)) as (n) approaches infinity does not approach zero, as it approaches negative infinity. Therefore, the alternating series test cannot be applied directly.

Another way to test for convergence is to split the series into two parts: one with even indices and one with odd indices, and then examine each part separately.

For even (n), (1-n^2) will be negative, and for odd (n), (1-n^2) will be positive. Therefore, the series can be rewritten as:

(\sum_{n=1}^{\infty} (-1)^n(1-n^2) = \sum_{n=1}^{\infty} -(1-n^2) + \sum_{n=1}^{\infty} (1-n^2))

Now, we can test the convergence of each part separately.

For the first part, (\sum_{n=1}^{\infty} -(1-n^2)), the terms are decreasing as (n) increases and the limit of the absolute value of the terms approaches zero as (n) approaches infinity. Therefore, by the alternating series test, this part converges.

For the second part, (\sum_{n=1}^{\infty} (1-n^2)), the terms are increasing as (n) increases and the limit of the absolute value of the terms does not approach zero as (n) approaches infinity. Therefore, this part diverges.

Since the series consists of alternating terms, and one part converges while the other diverges, the overall convergence of the series is inconclusive.

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