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

In order to determine if the series is convergent, we need to determine if the criteria are true.

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

The alternating series test states that if the terms of a series alternate in sign, decrease in absolute value, and have a limit of zero as ( n ) approaches infinity, then the series converges.

In this series, the terms alternate in sign (( (-1)^{n+1} )), and ( \frac{\ln(n+1)}{n+1} ) decreases as ( n ) increases because the natural logarithm function grows slower than any positive power of ( n ).

Furthermore, we can find the limit of ( \frac{\ln(n+1)}{n+1} ) as ( n ) approaches infinity:

[ \lim_{n \to \infty} \frac{\ln(n+1)}{n+1} = 0 ]

Since the limit of the terms of the series is zero, and the terms alternate in sign and decrease in absolute value, by the alternating series test, the series ( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \ln(n+1)}{n+1} ) converges.

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