How do I prove that the series ln(n)/(3n) is divergent?
To prove that the series ln(n)/(3n) is divergent, we can use the limit comparison test.
Here is how you can go about it:
1. Start by considering the series and write it in summation notation:
∑(n=1 to ∞) ln(n)/(3n)
2. Take the limit of the ratio of the general term of the given series and a known divergent series. In this case, we can choose the harmonic series ∑(n=1 to ∞) 1/n as our known divergent series.
Take the limit as n approaches infinity of ln(n)/(3n) / (1/n). Simplify and rearrange if needed.
3. Apply the limit comparison test. If the limit is greater than zero, the series diverges. If the limit is zero or infinity, the test is inconclusive.
By calculating the limit, you should obtain:
lim(n→∞) [ln(n)/(3n)] / (1/n) = lim(n→∞) ln(n)/(3n) * n/1 = lim(n→∞) (ln(n)/3)
4. Determine the limit of (ln(n)/3) as n approaches infinity.
By evaluating the limit, you will find:
lim(n→∞) (ln(n)/3) = ∞ (the limit is infinity)
5. Since the limit is not zero, we can conclude that the given series diverges by the limit comparison test.
Therefore, the series ln(n)/(3n) is divergent.