The ratio test states that a series #sum_(n=0)^oo a_n# converges absolutely if:
#lim_(n->oo) abs (a_(n+1)/a_n) < 1#
Let us determine the ratio for the series:
#sum_(n=0)^oo (4^n)/(3^n+5^n)x^n#
#abs (a_(n+1)/a_n) = (4^(n+1)/(3^(n+1)+5^(n+1))abs(x)^(n+1))/(4^n/(3^n+5^n)abs(x)^n)=4abs(x) (3^n+5^n)/(3^(n+1)+5^(n+1))=4/5abs(x) (1+(3/5)^n)/(1+(3/5)^(n+1))#
Now, as #3/5 < 1# we have that:
#lim_(n->oo) (3/5)^n = 0#
so that:
#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) 4/5abs(x) (1+(3/5)^n)/(1+(3/5)^(n+1)) = 4/5absx#
We can then conclude that for:
#absx <5/4 => lim_(n->oo) abs (a_(n+1)/a_n) <1# and the series is absolutely convergent
#absx > 5/4 => lim_(n->oo) abs (a_(n+1)/a_n) >1# and the series is divergent,
The case where #abs(x) = 5/4# is indeterminate and we have to analyze in detail.
In the case where #x=+-5/4# we have:
#abs(a_n) = 4^n/(3^n+5^n)(5/4)^n = 5^n/(3^n+5^n) = 1/(1+(3/5)^n)#
so that:
#lim_(n->oo) abs(a_n) = 1 >0#
and that means that the series cannot converge.
In conclusion the series is convergent in the interval #x in (-5/4,5/4)# where it is also absolutely convergent.
