Decide if ∑(n=2 to ∞) 1/ln(n!) converges or diverges.

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Understand the Problem

The question is asking to determine whether the infinite series starting from n=2 and summing the terms 1/ln(n!) converges or diverges. This involves analyzing the behavior of the series as n approaches infinity.

Answer

The series diverges.
Answer for screen readers

The series (\sum_{n=2}^{\infty} \frac{1}{\ln(n!)}) diverges.

Steps to Solve

  1. Analyze the term $\ln(n!)$

We will use Stirling's approximation, which states that for large $n$:

$$ \ln(n!) \sim n \ln(n) - n $$

This approximation helps us understand the growth rate of $\ln(n!)$ as $n$ becomes large.

  1. Rewrite the series term

Using Stirling's approximation, we can rewrite the series term:

$$ \frac{1}{\ln(n!)} \sim \frac{1}{n \ln(n) - n} $$

For large $n$, we can neglect the $-n$ in the denominator, simplifying our term to:

$$ \frac{1}{\ln(n!)} \sim \frac{1}{n \ln(n)} $$

  1. Determine the behavior of the series

Now, we need to analyze the convergence of the series:

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

We know that this series diverges by theIntegral Test or the Cauchy Condensation Test.

  1. Conclusion on convergence

Since

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

diverges, we conclude that

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

also diverges.

The series (\sum_{n=2}^{\infty} \frac{1}{\ln(n!)}) diverges.

More Information

The divergence of the series can be tied to the harmonic series behavior. The series involving (\frac{1}{n \ln(n)}) diverges, showing that (\ln(n!)) grows, but not rapidly enough to ensure convergence in the original series.

Tips

  • Failing to apply Stirling's approximation correctly or neglecting terms could lead to incorrect conclusions about divergence or convergence.
  • Not recognizing that the series (\sum_{n=2}^{\infty} \frac{1}{n \ln(n)}) diverges can mislead one into mistakenly believing that the original series might converge.

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