Why does the integral test not apply to #Sigma (-1)^n/n# from #[1,oo)#?

Answer 1
The integral test stipulates that for #suma_n# to apply, the terms of #a_n# must all be positive.
#(-1)^n/n# alternates between positive and negative terms.
You could use the integral test on #sum_(n=1)^oo1/n# since #1/n# is both positive and decreasing, which are the conditions for the integral test to apply.
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Answer 2

The integral test requires the function being integrated to be continuous, positive, and decreasing. The function ( \frac{(-1)^n}{n} ) oscillates in sign, so it fails to meet the criteria of being positive and decreasing. Therefore, the integral test cannot be applied to the series ( \sum_{n=1}^\infty \frac{(-1)^n}{n} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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