# How do you determine the convergence or divergence of #Sigma ((-1)^(n))/(ln(n+1))# from #[1,oo)#?

The series:

is convergent.

The series:

is an alternating series, so we can test its convergence using Leibniz's theorem, which states that an alternating series

is convergent if:

in our case:

also the second condition is satisfied.

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

- Check if the sequence ( \frac{1}{\ln(n+1)} ) is positive, decreasing, and approaches zero as ( n ) approaches infinity.
- Verify if it satisfies the conditions of the Alternating Series Test.

For the Alternating Series Test:

- The terms alternate in sign.
- The absolute value of the terms decreases as ( n ) increases.
- The terms approach zero as ( n ) approaches infinity.

- The sequence ( \frac{1}{\ln(n+1)} ) is positive for all ( n \geq 1 ).
- ( \frac{1}{\ln(n+1)} ) decreases as ( n ) increases because ( \ln(n+1) ) increases as ( n ) increases.
- As ( n ) approaches infinity, ( \frac{1}{\ln(n+1)} ) approaches zero.

Since the sequence ( \frac{1}{\ln(n+1)} ) satisfies all the conditions of the Alternating Series Test, the series ( \sum_{n=1}^{\infty} \frac{(-1)^n}{\ln(n+1)} ) 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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- How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #1/(1*2)+1/(2*3)+...+1/(n(n+1))+...#?
- What if L'hospital's rule doesn't work?
- How do you test for convergence of #Sigma (-1)^n(1-n^2)# from #n=[1,oo)#?

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