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)
_
Q
n = 1
3n
n +
This series can be written as:
∑ 1/ (3n/(n+1)) for n = 1 to infinity
To determine if the series converges or diverges, we can use the Direct Comparison Test.
Let's compare it to the series ∑ 1/n, which is a p-series with p = 1.
For the given series, we have:
1/ (3n/(n+1)) = (n+1)/(3n)
Now, we can compare this to 1/n:
(n+1)/(3n) < 1/n
To simplify, let's multiply through by n and rearrange:
n + 1 < 3n
Subtract n from both sides:
1 < 2n
Divide both sides by 2:
1/2 < n
Since n > 1/2 for all values of n in the series, we know that every term in the given series is less than the corresponding term in the series ∑ 1/n.
Since the series ∑ 1/n is a divergent p-series, and every term in the given series is less than the corresponding term in ∑ 1/n, we can conclude that the given series also diverges.
Therefore, the series diverges.