Determine whether the series converge or diverge

Question

I want to determine if $$\sum_{n=1}^\infty\frac{(-1)^{n-1}n}{1+n^\frac{4}{3}}\ln(1+\ln(n))$$ and $$\sum_{n=2}^\infty\frac{(\ln(n+1)-\ln(n))^a}{\ln^2(n)},a>1$$ converge or diverge. I have found that the ratio test and the root test are inconclusive.

Answer 1

For the first simply note that the series is an alternating series in the form

$$\sum (-1)^{n}a_n \quad a_n=\frac{n\ln(1+\ln n)}{1+n^\frac{4}{3}}$$

with $a_n$ decreasing and thus it converges.

For the second, as shown by Robert Z note that the "tail" is

$$\frac{(\ln(n+1)-\ln n)^a}{\ln^2 n}=\frac{(\ln(1+1/n))^a}{\ln^2 n}\sim\frac{1}{n^a\ln^2 n}$$

then apply the comparison test with $\sum b_n$ assuming

$$b_n=\frac{1}{n^a\ln^2 n}$$

Answer 2

Hint. For the first one, consider the function $$f(x)=\frac{x\ln(1+\ln(x))}{1+x^\frac{4}{3}}$$ and show that $f$ is positive for $x>1$, it is eventually decreasing (evaluate the limit of $x^{4/3}f'(x)$ as $x\to +\infty$), and $f(x)$ goes to zero as $x\to +\infty$. Then use Leibniz criterion.

As regards the second series, notice that $$\frac{(\ln(n+1)-\ln(n))^a}{\ln^2(n)}=\frac{(\ln(1+1/n))^a}{\ln^2(n)}\sim\frac{1}{n^a\ln^2(n)}.$$ Then use a comparison test.