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# Determine the convergence or divergence of the series.sum_{n=3}^inftyfrac{1}{n(ln n)[ln(ln n)]^4}

Question
Series
asked 2021-01-10

Determine the convergence or divergence of the series.
$$\sum_{n=3}^\infty\frac{1}{n(\ln n)[\ln(\ln n)]^4}$$

## Answers (1)

2021-01-11
A series, $$\sum_{n=m}^\infty f(n)$$ converges if the value of the integral, $$\int_m^\infty f(n)$$ is finite and diverges if it is infinite.
For the series, $$\sum_{n=3}^\infty\frac{1}{n(\ln n)[\ln(\ln n)]^4}$$, the integral is $$\int_3^\infty\frac{1}{x(\ln x)[\ln(\ln x)]^4}dx$$
Let $$\ln(\ln x)=y,\frac{1}{x\ln x}dx=dy$$
At $$x=3,y=\ln(\ln3)$$
at $$x=\infty,y=\infty$$
So, the integral becomes:
$$\int_3^\infty\frac{1}{x(\ln x)[\ln(\ln x)]^4}dx=\int_{\ln(\ln3)}^\infty\frac{1}{x(\ln x)(y)^4}x(\ln x)dy$$
$$=\int_{\ln(\ln3)}^\infty\frac{1}{y^4}dy$$
$$=-\frac{1}{5y^5}|_{\ln(\ln3)}^\infty$$
$$=-\frac15(\frac{1}{\infty}-\frac{1}{(\ln(\ln3))^5})$$
$$=\frac{1}{5(\ln(\ln3))^5}$$
Since, the value of the integral is finite. Thus, the sum of the series converges.

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