Calc - do these series converge or diverge?

Question

This is a question on last years version of an exam I'm studying for, can anyone help me out since my professor doesn't post or give answers?

Test each of the following series for convergence by either the Comparison Test or the Limit Comparison Test. If at least one 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 summation from n=1 to infinity:

1. ((ln(n))^6) / (n+5)
2. (9n^(8)-n^(4)+5n^(1/2)) / (5n^(10)-n^(3)+6)
3. (5n^(3)) / (n^(4)+3)
4. (cos(n)*n^(1/2)) / (5n+3)
5. (cos^(2)(n)*n^(1/2) / (n^(3))

Thanks!

Answer 1

1. Converges to 0. ln(n) increases more slowly than any positive power of n, and that would include the 1/6 or 1/12 power.

2. Converges to 0, because the biggest exponent in the denominator is bigger than the biggest exponent in the numerator.

3. See 2.

4. Bounded by 1 * n^(1/2)/(5n+3). From there, see 2.

5. Bounded by 1/n^3. From there, see 2.

One trick to recall is that if you see a sin or cos, its absolute value is less than or equal to 1, GUARANTEED.

Answer 2

i'd desire to show alternating series attempt on the 1st one... it shows convergence.. to ascertain no count selection selection if absolute or conditional... use assessment attempt... for the subsequent ... you need to stick to ratio attempt... by skill of fact the expression is composed of factorials... §