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

Question

8209 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-02T13:22:24

The series:

#sum_(n=1)^oo (-1)^n/ln(n+1)#

is convergent.

The series:

#sum_(n=1)^oo (-1)^n/ln(n+1)#

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

#sum_(n=1)^oo (-1)^n a_n#

is convergent if:

(i) #lim_(n->oo) a_n = 0#

(ii) #a_(n+1) <= a_n#

in our case:

#lim_(n->oo) 1/ln(n+1) = 0#

and since #lnx# is a monotone growing function:

#ln n < ln(n+1) <=> 1/lnn > 1/ln(n+1)#

also the second condition is satisfied.