Use the Ratio Test for Convergence/Divergence

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SUMMARY

The forum discussion focuses on applying the Ratio Test to determine the convergence or divergence of the series defined by the expression \( \sum_{k=1}^{\infty} \left(1 - \frac{1}{k}\right)^{3k} \). The analysis shows that the limit \( \lim_{k\to +\infty}\frac{u_{k+1}}{u_k} \) evaluates to 1, which is inconclusive. However, the term test reveals that \( \lim_{k\to +\infty} u_k = e^{-3} \neq 0 \), confirming that the series is divergent.

PREREQUISITES
  • Understanding of the Ratio Test for series convergence
  • Familiarity with limits and their properties
  • Knowledge of exponential functions and their limits
  • Basic concepts of series and sequences in calculus
NEXT STEPS
  • Study the application of the Root Test for series convergence
  • Explore the properties of exponential functions in calculus
  • Learn about the Divergence Test and its implications
  • Investigate other convergence tests such as the Comparison Test
USEFUL FOR

Students and educators in calculus, mathematicians analyzing series, and anyone interested in convergence tests for infinite series.

Fernando Revilla
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I quote a question from Yahoo! Answers

Use the ratio test to solve:? 1) k=1--> inf (1-(1/k))^(3k) 2)Is it convergent or divergent?

I have given a link to the topic there so the OP can see my response.
 
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We have

$\displaystyle\lim_{k\to +\infty}\frac{u_{k+1}}{u_k}=\lim_{k\to +\infty}\frac{\left(1-\frac{1}{k+1}\right)^{3k+3}}{\left(1-\frac{1}{k}\right)^{3k}}=\lim_{k\to +\infty}\left(1-\frac{1}{k+1}\right)^3\cdot\lim_{k\to +\infty}\frac{\left(1-\frac{1}{k+1}\right)^{3k}}{\left(1-\frac{1}{k}\right)^{3k}}$

$=1\cdot\displaystyle\lim_{k\to +\infty}\left(\frac{k^2}{k^2-1}\right)^{3k}=\ldots=1\cdot 1=1\text{ (inconclusive)}$

But $\displaystyle\lim_{k\to +\infty}u_k=\displaystyle\lim_{k\to +\infty}\left(1-\frac{1}{k}\right)^{3k}=\ldots=e^{-3}\ne 0$ so, (term test) the series is divergent.
 

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