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Help me explore the convergence of red color rounded series. On this photo (the equation below) I used radical indication but it doesn't show me the result. What would be better to use?

$$\color{red}{\sum_{n=1}^{\infty}\frac{1}{n^3\ln(n)}=\text{?}}$$ $$\lim_{n \to \infty}\sqrt[n]{\frac{1}{n^3\ln(n)}} = \lim_{n \to \infty}\frac{\sqrt[n]{1^n}}{\sqrt[n]{3^n} \sqrt[n]{\ln(n)}} = 1$$

Radical Cauchy indication does not help.

Doorknob
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Luchnik
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  • $$\left ( 3^n\right)^{1/n}=3$$ – Ron Gordon Apr 05 '14 at 11:30
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    First note that your series should start with $n=2$ (since $\ln(1)=0$) and in this case the series may be rewritten as $,\displaystyle\int_3^{\infty}(\zeta(x)-1),dx\approx 0.2379961002$. The computation is the same as here or here but with the lower bound $0$ (or $2$) replaced by $3$. – Raymond Manzoni Apr 05 '14 at 11:40

2 Answers2

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Use the comparison with the Riemann series $$\frac1{n^3\ln n}\le \frac1{n^3}\quad\text{for $n$ large enough}$$

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Since $\log (n) >1$ as soon as $n \geq 3$, you can write $$\sum _{n=2}^{\infty } \frac{1}{n^3 \log (n)}=\frac{1}{8 \log (2)}+\sum _{n=3}^{\infty } \frac{1}{n^3 \log (n)}$$ and now use the comparison suggested by Sami Ben Romdhane.

Added after Raymond Manzoni's comments

Just for entertainment, $$\sum _{n=3}^{\infty } \frac{1}{n^3}=\zeta (3)-\frac{9}{8}=0.077057$$ So,$$\sum _{n=2}^{\infty } \frac{1}{n^3 \log (n)}=0.237996 \le \frac{1}{8 \log (2)}+\zeta (3)-\frac{9}{8}=0.257394$$

Claude Leibovici
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  • @RaymondManzoni. Oooops ! – Claude Leibovici Apr 05 '14 at 11:50
  • I compare these two expresions as Sami Ben Romdhane suggested me. 1st is bigger than 2nd. When I take a limit(1st / 2nd) it equals 0. So what it's mean? – Luchnik Apr 05 '14 at 11:58
  • @Luchnik: the idea is that $\displaystyle \sum_{n=1} \frac 1{n^x}$ is convergent for $x>1$ (in your case $x=3$ and the previous sum is $\zeta(3)$). Your sum (starting with $n=2$) is positive and upper bounded by a convergent sum so... – Raymond Manzoni Apr 05 '14 at 12:02