Consider $$\sum \frac{1}{n\ \ln^{3/2}(n)}$$
The ratio test is inconclusive.
The root test is inconclusive.
And it seems right that $\frac{1}{n\ (\ln(n))^{3/2}}\leq\frac{1}{n}$ which diverges, but the correct answer is that the original sum does converge. I don't how to find a Majorant to it.
(i.e to apply a comparison test). Could any one help me?
Best Answer
In this problem, you should use the integration test of the Cauchy, which is stated as follows:
"Let $f:[k,+\infty)\to\mathbb{R}$ be a function satisfying the conditions: $f(x)>0$ for all $x\in [k,+\infty)$, and $f$ is decreasing function on $[k,+\infty)$. Then the series $\sum\limits_{n = k}^{ + \infty } {f\left( n \right)}$ is convergent if and only if the improper integral $\int\limits_k^{ + \infty } {f\left( x \right)dx}$ is convergent".
Solution: Define the function $f\left( x \right) = \frac{1}{{x{{\ln }^{3/2}}\left( x \right)}}$, $x\ge 2$. We see that $f(x)>0$ for all $x \ge 1$; and $f$ is the decreasing function on $[2,+\infty)$ since the derivative $$f'\left( x \right) = -\left( \frac{{{{\ln }^{3/2}}\left( x \right) + \frac{3} {2}{{\ln }^{1/2}}\left( x \right)}}{{{x^2}{{\ln }^3}\left( x \right)}}\right) < 0,\,\, \forall x\ge 2.$$ Therefore, using the integration test of the Cauchy as above, we conclude that the series $\sum\limits_{n = 2}^{ + \infty } {\frac{1}{{n{{\ln }^{3/2}}\left( n \right)}}}$ converges if and only if the integral $\int\limits_2^{ + \infty } {\frac{1}{{x{{\ln }^{3/2}}\left( x \right)}}dx}$ converges. But, we have $$\int\limits_2^{ + \infty } {\frac{1}{{x{{\ln }^{3/2}}\left( x \right)}}dx} = \int\limits_2^{ + \infty } {{{\left( {\ln x} \right)}^{ - \frac{3}{2}}}d\left( {\ln x} \right)} = \left. { - 2{{\left( {\ln x} \right)}^{ - \frac{1}{2}}}} \right|_{x = 2}^{x \to + \infty } = \frac{2}{{\sqrt {\ln 2} }}.$$ It means $\int\limits_2^{ + \infty } {\frac{1}{{x{{\ln }^{3/2}}\left( x \right)}}dx} $ converges. So, the series $\sum\limits_{n = 2}^{ + \infty } {\frac{1}{{n{{\ln }^{3/2}}\left( n \right)}}} $ is also convergent.