How do you apply the ratio test to determine if #Sigma (5^n)/(6^n-5^n)# from #n=[1,oo)# is convergent to divergent?

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Answer 1 · 2017-06-24T00:35:59

The series:

#sum_(n=1)^oo 5^n/(6^n-5^n)#

is convergent.

Evaluate the ratio:

#abs(a_(n+1)/a_n) = abs ( (5^(n+1)/(6^(n+1)-5^(n+1)))/(5^n/(6^n-5^n)))#

#abs(a_(n+1)/a_n) = (5^(n+1)/5^n ) ( (6^n-5^n) / (6^(n+1)-5^(n+1)))#

#abs(a_(n+1)/a_n) = 5*6^n( (1-(5/6)^n) / (6^(n+1)-5^(n+1)))#

#abs(a_(n+1)/a_n) = 5/6 *6^(n+1)( (1-(5/6)^n) / (6^(n+1)-5^(n+1)))#

#abs(a_(n+1)/a_n) = 5/6 ( (1-(5/6)^n) / ((6^(n+1)-5^(n+1) )/ 6^(n+1)))#

#abs(a_(n+1)/a_n) = 5/6 ( (1-(5/6)^n) / (1-(5/6)^(n+1) ))#

Now as #5/6 < 1#

#lim_(n->oo) (5/6)^n = lim_(n->oo) (5/6)^(n+1) = 0#

so:

#lim_(n->oo) abs(a_(n+1)/a_n) = 5/6 lim_(n->oo) ( (1-(5/6)^n) / (1-(5/6)^(n+1) )) = 5/6 < 1#

which proves the series to be convergent.