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

Question

2154 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-13T10:59:08

The alternating series:

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

is not convergent.

This is an alternating series so we can apply Leibniz's test stating that the series is convergent if given:

#a_n = 1/(root(n)(n)#

we have:

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

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

Now consider the sequence:

#b_n = ln a_n = ln (1/(root(n)(n))) = -1/n ln n#

We have:

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

And:

#lim_(n->oo) a_n =lim_(n->oo) e^(b_n)#

but as #e^x# is a continuous function:

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

The Leibniz's test is then not satisfied and the series is not convergent.