Determine the convergence or divergence of the series. ∑n=3∞1n(ln⁡n)[ln⁡(ln⁡n)]4

A series, ∑n=m∞f(n) converges if the value of the integral, ∫m∞f(n) is finite and diverges if it is infinite. For the series, ∑n=3∞1n(ln⁡n)[ln⁡(ln⁡n)]4, the integral is ∫3∞1x(ln⁡x)[ln⁡(ln⁡x)]4dx Let ln⁡(ln⁡x)=y,1xln⁡xdx=dy At x=3,y=ln⁡(ln⁡3) at x=∞,y=∞ So, the integral becomes: ∫3∞1x(ln⁡x)[ln⁡(ln⁡x)]4dx=∫ln⁡(ln⁡3)∞1x(ln⁡x)(y)4x(ln⁡x)dy =∫ln⁡(ln⁡3)∞1y4dy =−15y5|ln⁡(ln⁡3)∞ =−15(1∞−1(ln⁡(ln⁡3))5) =15(ln⁡(ln⁡3))5 Since, the value of the integral is finite. Thus, the sum of the series converges.

Explained: "Determine the convergence or divergence of the series...."

Ayaana Buck

Answered question

2021-01-10

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

Answer & Explanation

comentezq

Skilled2021-01-11Added 106 answers

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}} d x

Let

\ln (\ln x) = y, \frac{1}{x \ln x} d x = d y

x = 3, y = \ln (\ln 3)

x = \infty, y = \infty

So, the integral becomes:

\int_{3}^{\infty} \frac{1}{x (\ln x) [\ln (\ln x)]^{4}} d x = \int_{\ln (\ln 3)}^{\infty} \frac{1}{x (\ln x) (y)^{4}} x (\ln x) d y

= \int_{\ln (\ln 3)}^{\infty} \frac{1}{y^{4}} d y

= - \frac{1}{5 y^{5}} |_{\ln (\ln 3)}^{\infty}

= - \frac{1}{5} (\frac{1}{\infty} - \frac{1}{(\ln (\ln 3))^{5}})

= \frac{1}{5 (\ln (\ln 3))^{5}}

Since, the value of the integral is finite. Thus, the sum of the series converges.

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