# Studying the absolute and conditional convergence of $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^a(\ln(n))^3}$

$$\sum_{n=2}^{\infty} \frac{(-1)^n}{n^a(\ln n)^3}$$ I want to study the absolute convergence and conditional convergence of this series depending on $a$.

My work:

In class we have been using this convergence test:

$l= \lim_{x\to\infty} \frac{\ln(\frac{1}{a_n})}{\ln n}$ then if $l>1, a_n$ converges, if $l<1, a_n$ diverges and if $l=1$ it is unknown.

Using this and applying absolute value I have:

$$l= \lim_{n\to\infty}\frac{\ln(n^a(\ln n)^3)}{\ln(n)}=\frac{\ln(n^a)+\ln(\ln n )^3)}{\ln n}=\frac{a\ln n+3\ln(\ln n)}{\ln n}= \lim_{n\to\infty}a+\frac{3\ln(\ln n)}{\ln n}$$

$$\lim_{n\to\infty}a+\lim_{n\to\infty}\frac{3\ln(\ln n)}{\ln n}=a+0=a$$ This means that if $a<1$ the series diverges, if $a>1$ the series converges and if $a=1$ we dont know (and should study this case later)

If a series is absolutely converges it also converges, therefore the serie converges and absolutely converges for $a>1$. To finish the problem now I should study for which values of $0<a\leq{1}$ the series converges.

Is this correct? Are there any other easier convergence test I could use here?

• You meant $\sum_{n=2}^\infty$. Let $f_a(x) = \frac{1}{x^a \ln^3 x}$ note $f_a'(x) \sim \frac{c}{x^{a+1}\ln^3 x}$ and your series is $\sum_{n=1}^\infty \int_{2n}^{2n+1} f_a'(x)dx\sim \sum_{n=1}^\infty \frac{1}{2}\int_{2n}^{2n+2} f_a'(x)dx$. – reuns Oct 16 '17 at 15:21
• I'm not sure what comparison test you are using, but the alternating test shows your series can converge for some $a<1$. – Simply Beautiful Art Oct 16 '17 at 15:34