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everyone.

I have a doubt on the following question:

Let $ \left\{ \frac{a^n}{n!} \right\}, n \in \mathbb{N} $ be a sequence of real numbers, where $ a $ is a positive real number.

a) For what values of $ a $ is the sequence above monotonous? And bounded?

b) For what values of $ a $ is the sequence above convergent? Determine the limit of the sequence on that case.

How do I start this question?

Clearly 0 is a lower boundary. But is there a $ n_{max} $ after which $ a^n $ is always smaller than $ n! $ ?

As far as the monotonicity goes, I thought about splitting the problem into three cases:

$ \bullet $ For $ 0 < a < 1 $:

In this case, knowing that if $ 0 < a < 1 $, then $ a^{n+1} < a^n $, we can assume that $ A_{n+1} $ is always smaller than $ A_n $ because the numerator is decreasing (as showed) and the denominator is obviously increasing. Thus, if $ a_{n+1} < a_n, \forall \,\, n \in \mathbb{N} $, the sequence is strictly decreasing and so it is monotonic.

$ \bullet $ For $ a = 1 $:

In this case, the sequence is $ \frac{1}{n!} $ which is always positive, has 0 as a lower boundary and 1 as an upper boundary and thus it is bounded. Also, given that in this case $ A_{n+1} \leq A_n, \forall \,\, n \in \mathbb{N} $ the sequence is also monotonic.

$ \bullet $ For $ a > 1 $:

In this case, $ a^{n+1} > a^n, \forall \,\, n \in \mathbb{N} $. However, in order to known whether $ a_{n+1} < a_n $ or $ a_{n+1} > a_n $ we have to check if $ a^n > n! $, which I do not know how to do.

Could anyone help me with this question?


Thanks for the attention.

Kind regards, Pedro

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    $\begingroup$ Stirling's approximation might come quite handy in this particular situation: en.wikipedia.org/wiki/Stirling%27s_approximation $\endgroup$ – Victor Sep 10 '15 at 0:25
  • $\begingroup$ @Victor Thanks, mate. But this was a question of a test of Calculus II and the teacher never taught us that. This question was in the same test as questions like: determine whether the following series converges or diverges: $ \sum_{n=1}^{\infty} \left( \frac{5}{2^n} - \frac{1}{3^n} \right) $, so I really don't understand why this first question is so hard, given that the other aren't. $\endgroup$ – Pedro Cunha Sep 10 '15 at 0:30
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If $a_n = \frac{a^n}{n!} $, then $\frac{a_{n+1}}{a_n} =\frac{\frac{a^{n+1}}{(n+1)!}}{\frac{a^n}{n!}} =\frac{a}{n+1} $.

Therefore, if $a < n+1 $, then $a_n$ is decreasing. In particular, for any positive real $a$, $a_n$ is eventually monotonically decreasing.

If $n+1 < a$, then $a_n$ is increasing.

Therefore $a_n$ first increases and then decreases, with a peak at about $n=a$. Its value there is about (using Stirling) $\frac{n^n}{n!} \approx \frac{n^n}{\sqrt{2\pi n}\frac{n^n}{e^n}} = \frac{e^n}{\sqrt{2\pi n}} $.

Note that if $a=n$, then $a_n = a_{n+1}$ since $\frac{n^n}{n!} =\frac{n^{n+1}}{(n+1)!} $.

Many questions here show that $a_n \to 0$ as $n \to \infty$. You should be able to do this.

A more difficult question would be to get a more accurate estimate for the maximum term depending on how close $a$ is to an integer.

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  • $\begingroup$ Thanks for the answer, mate. It seems to me that the sequence will always be monotonic, because if $ a < n+1 $ the sequence is decreasing and thus it is monotonic and if $ a > n+1 $ the sequence is increasing and so the sequence is monotonic too. I didn't understand what you meant with: " Therefore $ a_n $ first increases and then decreases ". Won't it increase if $ a > n+1 $ and decrease if $ a < n+1 $ ? Both won't happen for the same value of $ a $ . Also, note that $ \frac{n^n}{n!} \neq \frac{n^{n+1}}{(n+1)!} $. $\endgroup$ – Pedro Cunha Sep 10 '15 at 2:39
  • $\begingroup$ rest of my comment: For b) I believe the first answer would be the interval where the sequence is both monotonic and bounded; the second answer is the result of $ \lim_{n \to \infty} \frac{n^n}{n!} $, which I believe I can answer on my own. However, I can't seem to understand how to find the answer to a). $\endgroup$ – Pedro Cunha Sep 10 '15 at 2:40
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    $\begingroup$ No. For any $a > 1$, the sequence will increase up until $n \sim a$ and then decrease from then on. The term for this is "unimodal". $\endgroup$ – marty cohen Sep 10 '15 at 2:42
  • $\begingroup$ I can't understand how do we go from $ a > n+1 $ and $ a < n+1 $ to $ a > 1 $. Are the ideas I proposed on my question correct for $ 0 < a \leq 1 $ ? It seems to me like it is and if they are, then the sequence is monotonic for $ 0 < a < 1 $ . $\endgroup$ – Pedro Cunha Sep 10 '15 at 2:49
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    $\begingroup$ For $a < 1$, the sequence is decreasing. But, for $a>1$, it is not. For example, for $a=3$ the sequence is 3, 9/2=4.5, 27/6=4.5, 81/24=3.375, ... which first increases then decreases. $\endgroup$ – marty cohen Sep 10 '15 at 3:04
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With the help of @marty cohen and @Victor I have conclude the following:

a)

For $ a < 1 $:

Since $ a^{n+1} < a^n $ if $ a < 1 $ and knowing also that $ \frac{1}{(n+1)!} < \frac{1}{n!}, \forall \,\, n \in \mathbb{N} $ it's easy for one to see that if $ a < 1 $, $ a_{n+1} < a_n, \forall \,\, n \in \mathbb{N} $ and thus the sequence will be strictly decreasing and as such it will be a monotonic sequence. Note also that $ \frac{a^n}{n!} > 0, \forall \,\, n \in \mathbb{N} $ and $ a_n < a_1, \forall \,\, n \in \mathbb{N} $, so the sequence is limited.

For $ a = 1$:

In this case the sequence can be re-written as $ \frac{1}{n!} $ which is strictly decreasing also given that $ \frac{1}{(n+1)!} < \frac{1}{n!}, \forall \,\, n \in \mathbb{N} $ . The sequence is bounded inferiorly by 0 and superiorly by 1 ( the smallest value $ \frac{1}{n!} $ can assume is 1 and the sequence is strictly decreasing). Being bounded both inferiorly and superiorly, the sequence is limited.

For $ a > 1 $:

In this case the sequence does not show a constant behavior. Taking $ a = 3 $, for example - credits to @marty cohen - we have:

$ a_1 = 3, a_2 = \frac{9}{2}, a_3 = \frac{9}{2}, a_4 = \frac{81}{24} = \frac{27}{8} < \frac{9}{2} $

which clearly shows that the sequence first increases and then decreases. Such sequence is called a unimodal sequence and it's not monotonic.

After examining the three cases, one can conclude that the answer is $ a \leq 1 $.

b)

By using the ratio-test, one can easily see the sequence is convergent because:

$ \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left | \frac{a}{n+1} \right| = 0, \forall \,\, a \in \mathbb{R^+} $

and so the convergence of the sequence is independent of the value one gives to $ a $.

Also, looking at $ \lim_{n \to \infty} \frac{a^n}{n!} $, one can clearly see this limit results in $ 0 $. This because:

$ \bullet a^n = \underbrace{a \cdot a \cdot a \dots a}_{n \text{ times} } $

$ \bullet n! \text{ as } n \to \infty = 1 \cdot 2 \cdot 3 \dots a \cdot (a+1) \cdot (a+2) \dots (a+k) \dots n, $ where $ k >> a $, and so, regardless of the value one gives to $ a $, $ n! $ will eventually outgrown $ a^n $.

Thank you, @marty cohen & @Victor for your help.

Kind regards, Pedro.

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Hint : $$\frac{a^n}{n!} = \frac{\underbrace{a \cdot a \cdot a... \cdot a}_{n \text{ times}}}{1\cdot 2 \cdot 3...\cdot n}$$

So for any finite positive parameter $a$ one can assess there are two numbers $n_1$ and $n_2$ such that $n_1 < a < n_2$, where both $n_1$, $n_2 \leq n$ as $n \rightarrow \infty$.

What can one conclude henceforth?

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  • $\begingroup$ Thanks for the answer. What I could understand from your answer was that for any given $ a $, we can affirm that exists two numbers $ n_1 $ and $ n_2 $ such that $ n_1 < a < n_2 $. That much I can see. What I can't see is how to use that to help find when $ a^n > n! $. Could you elaborate on your hint, please? $\endgroup$ – Pedro Cunha Sep 10 '15 at 2:00
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    $\begingroup$ In simpler terms, since $n$ gets infinitely large as the sequence is generated, any finite $a$ would ultimately be not enough to 'hold off' the factorial on the denominator, hence the sequence would $\rightarrow 0$ regardless of $a$'s value. This is merely an application of exponential growth versus factorial growth, considering $ n \rightarrow \infty $ . In light of these facts, one can easily find monotonicity by using the ratio test and prove convergence by finding the limit of the sequence, which will be obviously $0$ for any finite $a$. $\endgroup$ – Victor Sep 10 '15 at 2:09
  • $\begingroup$ Thought: if there is a $ n_1 $ such that for all $ a^k $ with $ k > n_1 $ we have $ a^k > k! $ or $ a^k < k! $ then for $ a > 1 $ the sequence is not monotonic, because if there's a change on it's behavior, we won't have $ a_n \leq a_{n+1} $ or $ a_n \geq a_{n+1} $, $ \forall \,\, n \in \mathbb{N} $ and thus the sequence won't be monotonic. Right? $\endgroup$ – Pedro Cunha Sep 10 '15 at 2:11
  • $\begingroup$ So for the sequence to be monotonic we would need to have $ \left| \frac{a}{n+1} \right| < 1 \Leftrightarrow -1 < \frac{a}{n+1} < 1 \Leftrightarrow -n-1 < a < n+1 $ ? $\endgroup$ – Pedro Cunha Sep 10 '15 at 2:15
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    $\begingroup$ No. You need to have $a_n \ge a_{n+1}$ for all $n$ since the series is eventually decreasing. $\endgroup$ – marty cohen Sep 10 '15 at 4:10

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