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

Dec 8, 2016

We will use the alternating series test, which says that for some series $\sum {\left(- 1\right)}^{n} {a}_{n}$, the series converges if ${a}_{n}$ is decreasing and ${\lim}_{n \rightarrow \infty} {a}_{n} = 0$.

We have the series ${\sum}_{n = 1}^{\infty} \frac{{\left(- 1\right)}^{n + 1} \ln \left(n + 1\right)}{n + 1}$. We see that ${\left(- 1\right)}^{n + 1}$ is the alternating portion so our sequence in question is ${a}_{n} = \ln \frac{n + 1}{n + 1}$.

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

We can see that $\ln \frac{n + 1}{n + 1}$ is decreasing by noting that $n + 1$ will increase faster than $\ln \left(n + 1\right)$, so the fraction will get smaller.

You could also graph this, or show that ${a}_{n - 1} / {a}_{n} < 1$, which means that the current term is greater than the previous term. If we want, we can show that $\frac{\ln \frac{n}{n}}{\ln \frac{n + 1}{n + 1}} < 1$.

The second criterion is that ${\lim}_{n \rightarrow \infty} \ln \frac{n + 1}{n + 1} = 0$. We can do this by again recognizing than $n$ grows far faster than $\ln \left(n\right)$.

We can also use L'Hôpital's rule to find the limit: ${\lim}_{n \rightarrow \infty} \ln \frac{n + 1}{n + 1} = {\lim}_{n \rightarrow \infty} \frac{\frac{1}{n + 1}}{1} = 0$.

Thus, ${\sum}_{n = 1}^{\infty} \frac{{\left(- 1\right)}^{n + 1} \ln \left(n + 1\right)}{n + 1}$ is a convergent series.