a) Divergent. As n becomes large, the term approaches 5/8. Because the sequence does not approach 0, the series is divergent.
b) Divergent. There are a number of ways to prove this; I will use the Limit Comparison Test, although the integral test is probably more intuitive.
lim as n→∞ of (log(n)/n)/(1/n) = lim as n→∞ log n = ∞
The harmonic series is divergent and the above shows that the terms in the series given grow at a greater rate than that. Therefore, it must be divergent.
Comments
a) Divergent. As n becomes large, the term approaches 5/8. Because the sequence does not approach 0, the series is divergent.
b) Divergent. There are a number of ways to prove this; I will use the Limit Comparison Test, although the integral test is probably more intuitive.
lim as n→∞ of (log(n)/n)/(1/n) = lim as n→∞ log n = ∞
The harmonic series is divergent and the above shows that the terms in the series given grow at a greater rate than that. Therefore, it must be divergent.
a) diverges because the sequence converges to 5/8 and not zero
b) diverges because the sequence's limit simplifies to 1/n which is a harmonic series and is always divergent