How do you test the alternating series #Sigma (-1)^n/lnn# from n is #[2,oo)# for convergence?

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Answer 1 · 2017-11-04T11:01:30

By the alternating series test criteria, the series converges

Suppose that we have a series #suma_n# and either

#a_n=(-1)^nb_n# or #a_n=(-1)^(n+1)b_n# where #b_n>=0# for all n.

Then if,

#1# #lim_(n->oo)b_n=0#

and,

#b_n# is a decreasing sequence

the series #suma_n# is convergent.

Here, we have

#sum_(n=2)^oo(-1)^n/lnn=sum_(n=2)^oo(-1)^n/lnn=sum_(n=2)^oo(-1)^n*1/lnn#

#b_n=1/lnn#

#lim_(n->oo)b_n=lim_(n->oo)(1/lnn)=0#

So, by the alternating series test criteria, the series converges