How do you test the series #Sigma (5^n+6^n)/(2^n+7^n)# from n is #[0,oo)# for convergence?

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Answer 1 · 2017-03-20T16:36:07

The series:

#sum_(n=0)^oo ((5^n+6^n)/(2^n+7^n))#

is convergent.

Use the ratio test by evaluating:

#abs(a_(n+1)/a_n) = ( ( (5^(n+1)+6^(n+1))/(2^(n+1)+7^(n+1)))/((5^n+6^n)/(2^n+7^n)))#

#abs(a_(n+1)/a_n) = ( (5^(n+1)+6^(n+1))/(2^(n+1)+7^(n+1))) ((2^n+7^n)/(5^n+6^n))#

#abs(a_(n+1)/a_n) = ( (5^(n+1)+6^(n+1))/(5^n+6^n)) ((2^n+7^n)/(2^(n+1)+7^(n+1)))#

#abs(a_(n+1)/a_n) = ( 6^(n+1)/6^n)( ((5/6)^(n+1)+1)/((5/6)^n+1)) (7^n/7^(n+1))(((2/7)^n+1)/((2/7)^(n+1)+1))#

#abs(a_(n+1)/a_n) = ( 6/7)( ((5/6)^(n+1)+1)/((5/6)^n+1)) (((2/7)^n+1)/((2/7)^(n+1)+1))#

We can now pass to the limit for #n->oo#:

#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo)( 6/7)( ((5/6)^(n+1)+1)/((5/6)^n+1)) (((2/7)^n+1)/((2/7)^(n+1)+1))#

Now as #5/6 < 1# we have:

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

and similarly:

#lim_(n->oo) (2/7)^n = lim_(n->oo) (2/7)^(n+1) = 0#

So:

#lim_(n->oo)( 6/7)( ((5/6)^(n+1)+1)/((5/6)^n+1)) (((2/7)^n+1)/((2/7)^(n+1)+1)) = 6/7 < 1#

which proves the series to be convergent.