Question 3. The numerical series
$$\sum^\infty_{n=1}(-1)^n\left(e^{1/(n+1)^5}-1\right)\log n$$
(A) converges but not absolutely
(B) is indeterminate
(C) positively diverges
(D) converges absolutely
I’ve been looking at this a while and I’m not really sure how to do this. I tried the leibniz test, but it’s not exactly trivial to show that this is monotone decreasing. So I moved on to looking for absolute convergence, however I can’t really think of a test that’d work for this except for comparison test. However I don’t really see a good candidate. Maybe $\ln{x}<x$ could be useful.
Best Answer
Hint: Consider absolute value and then use $e^{1/(n+1)^5}-1 \sim \frac{1}{(n+1)^5}, n\to \infty$.