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

Lipossig

Answered question

2020-12-21

Determine the convergence or divergence of the series.

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

Answer & Explanation

funblogC

Skilled2020-12-22Added 91 answers

Consider the given series as
$\sum_{n = 1}^{\infty} a_{n} = \sum_{n = 1}^{\infty} \ln (\frac{n + 1}{n})$
So, the nth term of the series is
$a_{n} = \ln (\frac{n + 1}{n})$
Consider another series $b_{n}$
$\sum_{n = 1}^{\infty} b_{n} = \sum_{n = 1}^{\infty} \ln (\frac{1}{n})$
where nth term of the series is $\ln (\frac{n + 1}{n})$ is greater than $\ln (\frac{1}{n})$
Therefore, the series $a_{n}$ is divergent series.
$b_{n} = \ln (\frac{1}{n})$
Use Comparison test for two series $a_{n}$ and $b_{n}$ such that $a_{n} > b_{n}$
$\frac{a_{n}}{b_{n}} = \frac{\ln (\frac{n + 1}{n})}{\ln (\frac{1}{n})} > 1$
Since, series $b_{n}$ is divergent series and $a_{n}$

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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