Ratio Test inconclusive with factorial

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SUMMARY

The discussion focuses on the convergence of the series \(\sum \frac{(n-1)!}{(n+2)!}\). The ratio test was applied, yielding a limit of \(\lim_{n \to \infty} \frac{n}{n+3} = 1\), which is inconclusive. To resolve this, the series was simplified to \(\frac{1}{(n+2)(n+1)n}\) and a comparison test was utilized with \(\frac{1}{n^3}\), demonstrating that the original series converges. For absolute convergence, it is essential to consider the absolute value of the original series.

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  • Understanding of series convergence tests, specifically the ratio test and comparison test.
  • Familiarity with factorial notation and simplification of series.
  • Knowledge of limits and their application in determining convergence.
  • Basic understanding of absolute and conditional convergence.
NEXT STEPS
  • Study the application of the Comparison Test in greater detail.
  • Learn about the Absolute Convergence Test and its implications.
  • Explore advanced series convergence tests, such as the Root Test.
  • Review factorial functions and their properties in series analysis.
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micast87
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Homework Statement



\sum(n-1)!/(n+2)!

2. The attempt at a solution

I tried the ratio test and came up with the lim_{}n\rightarrow\infty n/(n+3) = 1 which gives no information on convergence or divergence. I'm trying to find absolute or conditional convergence so what else can I do?
 
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You can first simplify that sum into 1/[(n+2)(n+1)(n)] = 1/[n^3 +3n^2 + 2n].

Then do comparison test using another sum: 1/n^3,

Since the latter is greater than the former and it converges, therefore the original series converges. For absolute convergence, just remember to take the absolute value of the original series.
 

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