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$$ \frac{2^n3^n}{n!} $$

I know the series is not monotonoic. I believe the series is bounded below at when n goes to infinity the series will go to 0, but never hit it. I just don't know if the series is bounded above.

Thomas Andrews
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Dank memer
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    This is a sequence. A series is a sum of (countably) infinitely many numbers. I have had a couple hundred students who ran afoul in an exam problem because they forgot the distinction. Hence the nitpick. Also a friendly piece of advice: making the distinction clear in your mind avoids misapplication of theorems, and unnecessary deductions from your exam scores! – Jyrki Lahtonen Apr 17 '17 at 06:12
  • The sequence is monotonic for n >6. – fleablood Apr 17 '17 at 06:54

2 Answers2

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$a_n = \dfrac{2^n3^n}{n!} = \dfrac{6^n}{n!}$

Then, $\displaystyle \lim_{n\to\infty}\dfrac{a_{n+1}}{a_n} = \lim_{n\to\infty}\dfrac{6}{n+1}=0$.

Therefore, the series converges, and is therefore bounded.

DHMO
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  • Much better if you say that since the series $\sum a_n$ is convergent (because of the Ratio Test), then $\lim_{n\to\infty} a_n=0$ and hence the sequence $(a_n)$ is bounded. – Juniven Acapulco Apr 17 '17 at 06:27
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Yes it's bounded. To see this, notice that it's monotonic decreasing for $n>6$.

Stella Biderman
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