How do you test the alternating series #Sigma (-1)^n(2^n+1)/(3^n-2)# from n is #[0,oo)# for convergence?

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Answer 1 · 2017-01-26T23:41:28

The series is convergent.

A sufficient condition for an alternating series to converge is established by the Leibniz test stating that if:

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

(ii) #a_(n+1) < a_n# for #n > N#

then the series is convergent.

In our case:

#lim_(n->oo) (2^n+1)/(3^n-2) = lim_(n->oo) ((2/3)^n -1/3^n)/(1-2/3^n) = 0#

so the first condition is satisfied.

Now consider:

#a_(n+1) = ((2^(n+1)+1)/(3^(n+1)-2)) = 2/3 ((2^n+1/2)/(3^n-2/3))#

clearly:

#(2^n+1/2) < (2^n+1)#

and:

#(3^n-2/3) > (3^n-2)#

so that:

#((2^n+1/2)/(3^n-2/3)) <((2^n+1)/(3^n-2))#

and it follows that:

#a_(n+1) = ((2^(n+1)+1)/(3^(n+1)-2)) = 2/3 ((2^n+1/2)/(3^n-2/3)) < 2/3((2^n+1)/(3^n-2)) < a_n#

Thus also the second condition is satisfied and the series os convergent.