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

Question

Christopher Allers · Accepted Answer

We will use the alternating series test, which says that for some series #sum(-1)^na_n#, the series converges if #a_n# is decreasing and #lim_(nrarroo)a_n=0#. We have the series #sum_(n=1)^oo((-1)^(n+1)ln(n+1))/(n+1)#. We see that #(-1)^(n+1)# is the alternating portion so our sequence in question is #a_n=ln(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(n+1)/(n+1)# is decreasing by noting that #n+1# will increase faster than #ln(n+1)#, 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 #(ln(n)/(n))/(ln(n+1)/(n+1))<1#. The second criterion is that #lim_(nrarroo)ln(n+1)/(n+1)=0#. We can do this by again recognizing than #n# grows far faster than #ln(n)#. We can also use L'Hôpital's rule to find the limit: #lim_(nrarroo)ln(n+1)/(n+1)=lim_(nrarroo)(1/(n+1))/1=0#. Thus, #sum_(n=1)^oo((-1)^(n+1)ln(n+1))/(n+1)# is a convergent series.

Charles Arocha · Answer

undefined 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.