∞∑k=11+ln(k)k
Determine whether it converges or diverges.
I don't think I could do limit comparison test because the ln(k) messed me up. Pretty sure I could do this with integral test but I think this is possible with comparison test as well. Could someone tell me if it is? For instance I'm looking for a bk value that is 0≤ak≤bk
My textbook uses bk=1k, but how is the hypothesis met with this? 1+ln(k)k≥1k
Thats wrong it should be ak≤bk ∀n≥1
Answer
The book is correct. Note that 1+log(k)≥1 for all k≥1. Hence,
1+log(k)k≥1k
Since the harmonic series diverges, then the series ∑∞k=11+log(k)k diverges by comparison. That is to say, the series of interest dominates the divergence harmonic series.
No comments:
Post a Comment