How do you determine if the series the converges conditionally, absolutely or diverges given #Sigma ((-1)^(n))/(lnn)# from #[1,oo)#?

Question

Cameron Alexander · Accepted Answer

The series is convergent.

We have that #a_n = ((-1)^n)/(log(n))# It is an alternating series.

This series is converget if

1) #lim_(n->oo) a_n = 0#
2) #abs(a_(n+1)) < abs(a_n)#

This series accomplishes both requirements so it is convergent.

Attached a plot of #S(n)=sum_(k=2)^n((-1)^k)/(logk)#

Emily Ammerman · Answer

undefined To determine the convergence of the series Σ((-1)^n)/(ln(n)) from n=1 to infinity, we can use the Alternating Series Test and the Ratio Test.

1. Alternating Series Test:
   - The series Σ((-1)^n)/(ln(n)) alternates in sign and has a decreasing absolute value as n increases.
   - Since lim(n→∞) (1/(ln(n))) = 0, the conditions of the Alternating Series Test are met.
   - Therefore, the series converges conditionally.

Therefore, the series Σ((-1)^n)/(ln(n)) converges conditionally.