#color(red)(sum_(n=1)^∞ (2x^4)/(x^5-10)" is divergent")#.
#sum_(n=1)^∞ (2x^4)/(x^5+10)#
The limit comparison test states that if #a_n# and #b_n# are series with positive terms and if #lim_(n→∞) (a_n)/(b_n)# is positive and finite, then either both series converge or both diverge.
Let #a_n = (2x^4)/(x^5+10)#
Let's think about the end behaviour of #a_n#.
For large #n#, the denominator #x^5+10# acts like #x^5#.
So, for large #n#, #a_n# acts like #(2x^4)/x^5 = 2/x#.
Let #b_n= 1/x#
Then #lim_(n→∞)(a_n/b_n) = lim_(n→∞)( ((2x^4)/(x^5+10))/(1/x)) = lim_(n→∞)( (2x^4×x)/(x^5+10)) = lim_(n→∞)( (2x^5)/(x^5+10)) = lim_(n→∞)( 2/(1-1/x^5)) =2#
The limit is both positive and finite, so either #a_n# and #b_n# are both divergent or both are convergent.
But #b_n= 1/x# is divergent, so
#a_n = x^4/(x^5-10)# is also divergent.
