Determine if the series converges or diverges.. Choose your convergence test. Choose from the following:integral,direct

Question

comparison,p-series, nth term test(aka divergence test), and ratio test. 5)
_
Q
n = 1
3n
n + 7
5) 6)
_
Q
n = 1
4n!
n
n
6) 7)
_
Q
n=1
7
n
7) 8)
_
Q
n=3
1
2n -5

Answer 1

For the series 1/3n/(n+7), we can use the direct comparison test.

Divide the numerator and denominator of the series by n. This yields:

1/(3n/(n+7))

Now, we can compare the series to 1/n.

As n approaches infinity, the term 1/3n/(n+7) approaches 1/n. Since the series 1/n is a harmonic series and diverges, the original series also diverges by the direct comparison test.

For the series 4n!/(n^n), we can use the ratio test.

Taking the ratio of consecutive terms, we have:

(n+1)!/(n+1)^(n+1) * n^n / n!

The n! terms cancel out:

(n+1)/(n+1)^(n+1) * n^n

Simplify:

1/(n+1) * n^n

As n approaches infinity, the ratio of consecutive terms approaches 0. Therefore, the series converges by the ratio test.

For the series 7/n, we can use the p-series test.

The series is of the form 1/n^p, where p = 1. Since p is less than or equal to 1, the series diverges by the p-series test.

For the series 1/(2n-5), we can use the direct comparison test.

We can compare the series to 1/n.

As n approaches infinity, the term 1/(2n-5) approaches 1/n. Since the series 1/n is a harmonic series and diverges, the original series also diverges by the direct comparison test.