Calculate $\sum_{n=2}^{\infty} \frac{1}{n\ln(n)^2}$

Asked Nov 14 '21 at 22:02

Active Nov 14 '21 at 22:21

Viewed 72 times

Calculate $\sum_{n=2}^{\infty} \frac{1}{n\ln(n)^2}$ by direct addition to 5 decimal places.

I tried to use the following: $$\int_{N+1}^{\infty} f(x)dx \le \sum_{n=N+1} a_n \le \int_{N+1} ^{\infty} f(x)dx + a_{N+1} $$

Though I'm not sure if i'm plugging the numbers in correctlty;

$$\int_{2}^{\infty}\frac{1}{x\ln(x)^2} dx\le \sum_{2}^{\infty}\frac{1}{n\ln(n)^2} \le \left(\int_2^{\infty}\frac{1}{x\ln(x)^2}dx\right) +\frac{1}{n\ln(n)^2}$$

$$\implies -\frac{1}{\ln(2)} \le \sum_{2}^{\infty}\frac{1}{n\ln(n)^2} \le-\frac{1}{\ln(2)} + \frac{1}{n\ln(n)^2}$$

I would appreciate some guidance on how to proceed from here!

edited Nov 14 '21 at 22:21

asked Nov 14 '21 at 22:02

me.limes

1

I think the $\frac1{n\ln(n)}$ term should be $\frac1{2\ln(2)^2}$, since you are summing the first inequality from $N=1$ to $\infty$. If instead you were summnig from $N$ (arbitrary) to $\infty$, you would get $$\sum_{n=N+1}^\infty\frac1{n\ln(n)^2}\le\frac1{\ln(N+1)}+\frac1{(N+1)\ln(N+1)^2},$$ and you can make the right-hand side smaller than $10^{-5}$ by taking $N$ sufficiently large. – nejimban Nov 14 '21 at 22:15
1

Does this answer your question? On convergence of Bertrand series $\sum\limits_{n=2}^{\infty} \frac{1}{n^{\alpha}\ln^{\beta}(n)}$ where $\alpha, \beta \in \mathbb{R}$ – Matcha Latte Nov 21 '21 at 21:31

Calculate $\sum_{n=2}^{\infty} \frac{1}{n\ln(n)^2}$

0 Answers0