Factorial Series Convergence: Investigating the Sum of n!/10^n

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SUMMARY

The series \(\sum \frac{n!}{10^n}\) diverges as established through the ratio test. The ratio \(r = \frac{n+1}{10}\) indicates divergence for large \(n\) since \(r\) approaches infinity as \(n\) increases. Although the series appears convergent for finite \(n\) values (1 to 8), the behavior for large \(n\) is the determining factor. Thus, the ratio test is applicable and confirms divergence.

PREREQUISITES
  • Understanding of factorial notation and its properties.
  • Familiarity with convergence tests, specifically the ratio test.
  • Basic knowledge of infinite series and their behavior.
  • Proficiency in mathematical notation, including summation and limits.
NEXT STEPS
  • Study the implications of the ratio test in greater depth.
  • Explore other convergence tests such as the root test and comparison test.
  • Investigate the behavior of series involving factorials and exponential functions.
  • Learn about the convergence of power series and their applications.
USEFUL FOR

Students studying calculus, particularly those focusing on series convergence, mathematicians analyzing factorial growth, and educators teaching convergence tests in mathematical analysis.

Joshk80k
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Homework Statement



Show that

\sum \frac{n!}{10^n}

converges or diverges.(Note, I was unsure of how to format this via latex, so the summation is from n = 1 to infinity.)

Homework Equations



The root test:

|\frac{a_n_+_1}{a_n}|

The Attempt at a Solution



a_n=\frac{n!}{10^n},<br /> <br /> a_n_+_1\frac{(n+1)!}{10^n^+^1} = \frac{(n+1)n!}{10^n^+^1}

Applying the ratio test,

|\frac{a_n_+_1}{a_n}|= \frac{10^n(n+1)n!}{n!10^n^+^1}

Cancelling terms out,

\frac{n+1}{10} = r

Now, I know that if:

r > 1, it is divergent,
r < 1, convergent,
r = 1, inconclusive.

My problem is that I am not sure where to go after this. I still have an "n" in my answer, and I expected to just have a numerical answer.

I was going to just go ahead and say that since n approaches infinity, r is greater than 1, and thus the series is divergent, but I stopped because I realized that from n = 1 to n = 8, the series would be convergent, and worse still, at n = 9, the test would be inconclusive?

So my question is, did I make a mistake somewhere here, or is the ratio test not applicable here for this reason?

Thanks for any feedback!
 
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Joshk80k said:

Homework Statement



Show that

\sum \frac{n!}{10^n}

(Note, I was unsure of how to format this via latex, so the summation is from n = 1 to infinity.)

converges or diverges.



Homework Equations



The root test:

|\frac{a_n_+_1}{a_n}|

The Attempt at a Solution



a_n=\frac{n!}{10^n},<br /> <br /> a_n_+_1\frac{(n+1)!}{10^n^+^1} = \frac{(n+1)n!}{10^n^+^1}

Applying the ratio test,

|\frac{a_n_+_1}{a_n}|= \frac{10^n(n+1)n!}{n!10^n^+^1}

Cancelling terms out,

\frac{n+1}{10} = r

Now, I know that if:

r > 1, it is divergent,
r < 1, convergent,
r = 1, inconclusive.

My problem is that I am not sure where to go after this. I still have an "n" in my answer, and I expected to just have a numerical answer.

I was going to just go ahead and say that since n approaches infinity, r is greater than 1, and thus the series is divergent, but I stopped because I realized that from n = 1 to n = 8, the series would be convergent, and worse still, at n = 9, the test would be inconclusive?
It doesn't matter what happens between 1 and 8; you're interested in the behavior for large n. Every finite series converges, since you're just adding a finite number of terms.
Joshk80k said:
So my question is, did I make a mistake somewhere here, or is the ratio test not applicable here for this reason?

Thanks for any feedback!
 
Alright, so I was right - it diverges.

Thanks very much!
 

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