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

Test each of the following series for convergence by either the Comparison Test or the Limit Comparison Test. If either test can be applied to the series, enter CONV if it converges or DIV if it diverges. If neither test can be applied to the series, enter NA. (Note: this means that even if you know a given series converges by some other test, but the comparison tests cannot be applied to it, then you must enter NA rather than CONV.)

all are the sum from n = 1 to infinity
a) [(3n^9)-(n^6)+(8sqrt(n))]/[(8n^11)-(n^5)…
b) [(cos(n))(sqrt(n))]/[8n+7]
c) [(-1)^n]/[8n]
d) [(cos^2(n))(sqrt(n))]/[n^5]
e) [8(n^5)]/[(n^6)+7]

I got CONV for a) and e) and NA for the rest but one or more is incorrect.

a) Compare to Σ n^9/n^11 = Σ 1/n^2 (which a convergent p-series).

Using the Limit Comparison Test,
lim(n→∞) [(3n^9 - n^6 + 8 sqrt(n)) / (8n^11 - n^5)] / (1/n^2) = 3/8.

Hence, the series in question also converges.

b) NA, since the terms can be negative, due to the cosine term.

c) NA, since half the terms are negative.

d) Use the Comparison Test (since cos^2(n) > 0).
cos^2(n) n^(1/2) / n^5 < n^(1/2) / n^5 = 1/n^(9/2).

Since Σ 1/n^(9/2) is a convergent p-series, the original series also converges.

e) Use the Comparison Test.
8n^5/(n^6 + 7) > 8n^5/(n^6 + 7n^6) = 1/n for all n > 0.

Since Σ 1/n is the divergent harmonic series, the original series also diverges.

I hope this helps!