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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?

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    $\begingroup$ You might have more luck with the integral test or by looking at the sequence of partial sums. $\endgroup$
    – AlexR
    Commented Sep 12, 2014 at 11:53
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    $\begingroup$ It is crucial what you mean by $\ln(n)^{3/2}$. If it is $n(\ln n)^{3/2}$ in the denominator, it converges. $\endgroup$
    – amWhy
    Commented Sep 12, 2014 at 11:54
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    $\begingroup$ $\ln^{3/2}(n) \neq \ln(n^{3/2})$. Any sum with general term $$\frac 1{n\ln^p(n)}$$ converges for $p > 1$, else, diverges. $\endgroup$
    – amWhy
    Commented Sep 12, 2014 at 11:59
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    $\begingroup$ Dear Udi use the integral test then it is convergent $\endgroup$
    – M.R. Yegan
    Commented Sep 12, 2014 at 12:03
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    $\begingroup$ Udi: if you have Baby Rudin, see Thm. 3.29, p. 62, which uses Thm 3.27 for justification. $\endgroup$
    – amWhy
    Commented Sep 12, 2014 at 12:07

4 Answers 4

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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.

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  • $\begingroup$ That's great. I must thank you a lot. $\endgroup$
    – E Be
    Commented Sep 12, 2014 at 12:40
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Given $$ \int_2^\infty{\frac{dx}{x \log(x)^{3/2}}} $$ set $u = \log(x)$; then $du = dx/x$ and the integral is $$ \int_{\log(2)}^\infty \frac{du}{u^{3/2}} $$ which converges.

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The result comes from the generalised harmonic series:

$\sum_{n=2}^\infty \frac{1}{n\ ^\alpha(\ln(n))^{\beta}}$ $\begin{cases}\text{converges if } \alpha>1\text{ or if } \alpha=1 \text{ and } \beta>1 \\\text{diverges if }\alpha<1\text{ or if }\alpha=1\text{ and }\beta≤1 \end{cases}$

My professor stated this at a lecture but I do not have a proof

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Since the sequence is positive and decreasing we can apply the Cauchy condensation test $\sum_n a_n$ converges if and only if $\sum_n 2^n \cdot a_{2^n}$ converges. The condensed series is $$\sum_n \frac{2^n}{2^n \cdot (\log 2^n)^{3/2}}= \sum_n \frac{1}{(n \log 2)^{3/2}}$$ which is convergent (condense again if in doubt) so yes, the series is convergent.

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