Determine if the series converges or diverges.. Choose your convergence test. Choose from the following:integral,direct
comparison,p-series, nth term test(aka divergence test), and ratio test. 5)
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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
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.