Test each of the following series for convergence by either the Comparison Test or the Limit Comparison Test?

Question

If you can apply either the comparison test or limit comparison test then tell wheter it is convergent or divergent, if neither test can be applied to the series than say what test does
summation n=1 to infinity (cos(n)sqrt(n))/(9n+3)
summation n=1 to infinity ((ln(n))^2)/(n+9)
summation n=1 to infinity (9n^5-n^3+6n^(1/2))/(2n^7-n^2+2)
summation n=1 to infinity (9n^6)/(n^9+1)

Answer 1

1) tests can't be used for this series as its terms are not of constant sign, ie. cos 2, cos 5 are negative while cos 1 is positive
2) compare with 1/(n+9) since ln n > 1 if n>3 thus D
3) compare with 9/(2 n^2) using LCT thus C
4) compare with 9/(n^3) using LCT thus C

Answer 2

For 9n^5/(n^9 + 1), compare with 1/n^3. The series converges absolutely.

Answer 3

to your first question, sign is generally irrelevant, as long because it relatively is finite: lim a_n/(-b_n) is then advantageous, so sum a_n and sum (-b_n) (and for that reason sum b_n) converge or diverge together. on your 2d question, you may desire to be good. There exist sequence sum a_n and sum b_n that fulfill your standards with sum a_n convergent, and there exist sequence sum a_n' and sum b_n' that fulfill your standards with sum a_n' divergent. case in point, do no longer forget approximately a_n' = a million/n and b_n' = a million/n^2. Then a_n'/b_n' = n -> infinity and sum b_n' converges, yet sum a_n' diverges. even with the certainty that, do no longer forget approximately a_n = (-a million)^n/n and b_n = (-a million)^n/n^2. even with the certainty that a_n/b_n = n -> infinity and sum b_n converges, and sum a_n CONVERGES.