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

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.
0

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