If #sum_k a_n = sum_k (-2n^4 + 5n^2 - 3)/(n^11 - 9n^3 - 9)# and #sum_k b_n = sum_k 1/n^beta#, then for what value of #beta# could allow #a_n/b_n# to converge?

I got #beta = 7#. This uses the idea that at large #n#, polynomial limits will allow you to cross out lower-order terms.

The limit comparison test first assumes we have two sums, #sum a_n# and #sum b_n#, where #a_n >=0# and #b_n > 0#.

If

#lim_(n->oo) a_n/b_n = c#,

where #c# is a finite constant,

then both series converge or both diverge. In this case, we choose (and I think your infinite sum has a typo... #k# should be #n#):

#a_n = (-2n^4 + 5n^2 - 3)/(n^11 - 9n^3 - 9)#

#b_n = 1/n^beta#

What we're saying here is that if...

#c = lim_(n->oo) ((-2n^4 + 5n^2 - 3)/(n^11 - 9n^3 - 9))/(1//n^beta)#

#= lim_(n->oo) (-2n^4 + 5n^2 - 3)/(n^11 - 9n^3 - 9) cdot n^(beta)#,

then the series converges for the appropriate exponent #beta#. For #a_n#, as #n->oo#, the terms with higher exponents get larger more quickly than those with lower exponents.

So, in other words:

#lim_(n->oo) (-2n^4 + 5n^2 - 3)/(n^11 - 9n^3 - 9) = lim_(n->oo) -(2n^4)/(n^11)#

Therefore, for large #n#, we can rewrite this limit as:

#c = lim_(n->oo) (-2n^4)/(n^11) cdot n^(beta)#

#= lim_(n->oo) (-2)/(n^7) cdot n^(beta)#

Since #c# must be a finite constant, the easiest way to satisfy this is to have #color(blue)(beta = 7)#, as constants do not depend on the value of #n#.

#=> c = -lim_(n->oo) 2/(cancel(n^7)) cdot cancel(n^7)#

#= -2 = "finite constant"#