# How do you determine the convergence or divergence of #sum_(n=1)^(oo) (-1)^(n+1)/n#?

The series is convergent and its sum is:

Leibniz's theorem states that a sufficient condition for the series with alternating signs:

In fact:

and we can easily conclude that:

so that the MacLaurin series is:

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To determine the convergence or divergence of the series ( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} ), we can use the Alternating Series Test:

- Verify that the sequence ( \frac{1}{n} ) is positive, decreasing, and approaches zero as ( n ) approaches infinity.
- Confirm that the terms alternate in sign.

Since both conditions are met, the Alternating Series Test states that if a series ( \sum_{n=1}^{\infty} (-1)^{n+1} a_n ) satisfies the following conditions:

- ( a_n ) is positive, decreasing, and approaches zero as ( n ) approaches infinity.
- The terms alternate in sign.

Then the series converges.

Therefore, the series ( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} ) converges.

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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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