Determine the convergence or divergence of the series. ∑n=2∞1n(ln⁡n)3

Consider the series ∑n=2∞1n(ln⁡n)3 Cauchy Condensation test: Let () be a decreasing sequence of positive real numbers. Then the two series ∑n=1∞cn and ∑n=1∞2nc2n either convergent or divergent. Here ∑n=2∞1n(ln⁡n)3 (=frac{1}{n(ln n)^3}) monotonically decreasing series of positive real numbers for n≥2 ∑n=2∞cn=∑n=2∞1n(ln⁡n)3 ∑n=2∞2nc2n=∑n=2∞2n12n(ln⁡2n)3=∑n=2∞1(nln⁡2)3=∑n=2∞1(ln⁡2)3n3 ∑n=2∞1n3 is convergent by p-series test ⇒∑n=1∞2nc2n=1(ln⁡2)3∑n=2∞1n3 is convergent Hence by Cauchy Condensation test ∑n=1∞cn and ∑n=1∞2nc2n converge or diverge together. Here ∑n=1∞2nc2n=1(ln⁡2)3∑n=2∞1n3 convergent implies ∑n=2∞cn=∑n=2∞1n(ln⁡n)3 is convergent. Answer: Convergent

Check the solution to "Determine the convergence or divergence of the series...."

vestirme4

Answered question

2020-11-06

Determine the convergence or divergence of the series.

\sum_{n = 2}^{\infty} \frac{1}{n (\ln n)^{3}}

Answer & Explanation

Willie

Skilled2020-11-07Added 95 answers

Consider the series

\sum_{n = 2}^{\infty} \frac{1}{n (\ln n)^{3}}

Cauchy Condensation test:
Let \(\) be a decreasing sequence of positive real numbers. Then the two series

\sum_{n = 1}^{\infty} c_{n}

and

\sum_{n = 1}^{\infty} 2^{n} c_{2^{n}}

either convergent or divergent.
Here

\sum_{n = 2}^{\infty} \frac{1}{n (\ln n)^{3}}

\(=\frac{1}{n(\ln n)^3}\) monotonically decreasing series of positive real numbers for

n \geq 2

\sum_{n = 2}^{\infty} c_{n} = \sum_{n = 2}^{\infty} \frac{1}{n (\ln n)^{3}}

\sum_{n = 2}^{\infty} 2^{n} c_{2^{n}} = \sum_{n = 2}^{\infty} 2^{n} \frac{1}{2^{n} (\ln 2^{n})^{3}} = \sum_{n = 2}^{\infty} \frac{1}{(n \ln 2)^{3}} = \sum_{n = 2}^{\infty} \frac{1}{(\ln 2)^{3} n^{3}}

\sum_{n = 2}^{\infty} \frac{1}{n^{3}}

is convergent by p-series test

\Rightarrow \sum_{n = 1}^{\infty} 2^{n} c_{2^{n}} = \frac{1}{(\ln 2)^{3}} \sum_{n = 2}^{\infty} \frac{1}{n^{3}}

is convergent
Hence by Cauchy Condensation test

\sum_{n = 1}^{\infty} c_{n}

and

\sum_{n = 1}^{\infty} 2^{n} c_{2^{n}}

converge or diverge together.
Here

\sum_{n = 1}^{\infty} 2^{n} c_{2^{n}} = \frac{1}{(\ln 2)^{3}} \sum_{n = 2}^{\infty} \frac{1}{n^{3}}

convergent implies

\sum_{n = 2}^{\infty} c_{n} = \sum_{n = 2}^{\infty} \frac{1}{n (\ln n)^{3}}

is convergent.
Answer: Convergent

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Didn't find what you were looking for?