# How to check monotony of sequence

• MHB
• evinda
In summary: Thinking)In summary, we discussed the sequence $a_n=\frac{8^n}{n!}$, and found that it is not monotonic due to the alternating sign of the difference between consecutive terms. We also explored the possibility of using other criteria to determine monotonicity, but were unable to find one. We then moved on to discussing another sequence, $d_n=\frac{n^{30}}{2^n}$, and found that it is eventually decreasing. We also determined that the sequence is bounded below by $0$ and above by the integer part of $\frac{1}{2^{\frac{1}{30}}-1}$. However, this value does not represent a term in the sequence, but rather marks
evinda
Gold Member
MHB
Hello! (Wave)

Consider the sequence $a_n=\frac{8^n}{n!}$.

It holds that

$$a_{n+1}-a_n=\frac{8^{n+1}}{(n+1)!}-\frac{8^n}{n!}=\frac{8 \cdot 8^n}{(n+1) \cdot n!}-\frac{8^n}{n!}=\frac{8^n}{n!}\left( \frac{8}{n+1}-1\right)=\frac{8^n}{n!} \left( \frac{7-n}{n+1}\right)$$

Since the last term is positive for some $n$ and negative for others, we cannot conclude like that if the sequece is monotonic or not.

I haven't thought of an other criterion which we could use to see if the sequence is monotonic or not. Could you give me a hint? (Thinking)

evinda said:
Hello! (Wave)

Consider the sequence $a_n=\frac{8^n}{n!}$.

It holds that

$$a_{n+1}-a_n=\frac{8^{n+1}}{(n+1)!}-\frac{8^n}{n!}=\frac{8 \cdot 8^n}{(n+1) \cdot n!}-\frac{8^n}{n!}=\frac{8^n}{n!}\left( \frac{8}{n+1}-1\right)=\frac{8^n}{n!} \left( \frac{7-n}{n+1}\right)$$

Since the last term is positive for some $n$ and negative for others, we cannot conclude like that if the sequece is monotonic or not.

I haven't thought of an other criterion which we could use to see if the sequence is monotonic or not. Could you give me a hint? (Thinking)

Hey evinda!

Haven't you just shown that it is not monotonic? (Worried)

For $n<7$ the difference is positive, so the sequence is ascending.
At $n=7$ we find that $a_{7+1}-a_7=0$.
And for $n>7$ the sequence is descending. (Thinking)

Klaas van Aarsen said:
Hey evinda!

Haven't you just shown that it is not monotonic? (Worried)

For $n<7$ the difference is positive, so the sequence is ascending.
At $n=7$ we find that $a_{7+1}-a_7=0$.
And for $n>7$ the sequence is descending. (Thinking)
And the sequence is ascending for the interval $(0,7]$ and descending for the interval $(7,+\infty)$, right? (Thinking)If we consider the sequence $d_n=\frac{n^{30}}{2^n}$, we know that it is not ascending, since $d_1=\frac{1}{2}$ and $\lim_{n \to +\infty} d_n=0$.

So it could be only descending from some point and on.

Then it should hold $d_{n+1}-d_n \frac{(n+1)^{30}-2n^{30}}{2^{n+1}}<0$.

This holds for $\log_{2}\left( 1+\frac{1}{n}\right) \leq \frac{1}{30}$.

Can this be solved or to we let it as it is? (Thinking)

Klaas van Aarsen said:
Hey evinda!

Haven't you just shown that it is not monotonic? (Worried)

For $n<7$ the difference is positive, so the sequence is ascending.
At $n=7$ we find that $a_{7+1}-a_7=0$.
And for $n>7$ the sequence is descending. (Thinking)

Do we use $n=7$ to find the under or upper bound of the sequence? (Thinking)

evinda said:
And the sequence is ascending for the interval $(0,7]$ and descending for the interval $(7,+\infty)$, right?

Yep. (Nod)

evinda said:
If we consider the sequence $d_n=\frac{n^{30}}{2^n}$, we know that it is not ascending, since $d_1=\frac{1}{2}$ and $\lim_{n \to +\infty} d_n=0$.

So it could be only descending from some point and on.

Then it should hold $d_{n+1}-d_n \frac{(n+1)^{30}-2n^{30}}{2^{n+1}}<0$.

This holds for $\log_{2}\left( 1+\frac{1}{n}\right) \leq \frac{1}{30}$.

Can this be solved or to we let it as it is?

We can also write it as:
$$1+\frac 1n<2^{1/30}$$
can't we?
Can we find n now? (Wondering)

evinda said:
Do we use $n=7$ to find the under or upper bound of the sequence? (Thinking)

Yep. We found that $a_7=a_8$ is the maximum of the sequence. (Emo)

We can say that the sequence is eventually decreasing, meaning that, after a certain point (here, n= 7), it becomes decreasing.

Klaas van Aarsen said:
Yep. (Nod)
We can also write it as:
$$1+\frac 1n<2^{1/30}$$
can't we?
Can we find n now? (Wondering)

First of all, it should be $d_{n+1}-d_n \geq 0$, right?

This holds for $\left( 1+\frac{1}{n}\right)^{30} \geq 2$. Do we get from this that $1+\frac{1}{n} \geq 2^{\frac{1}{30}}$ although $30$ is even?

Then we would get that $n \leq \frac{1}{2^{\frac{1}{30}}-1}$.

So does this mean that the sequence hat its maximum value at the largest integer smaller than $\frac{1}{2^{\frac{1}{30}}-1}$ ? (Thinking)

Klaas van Aarsen said:
Yep. (Nod)

Yep. We found that $a_7=a_8$ is the maximum of the sequence. (Emo)

Since the sequence is descending for $n>7$, we have that $a_n< a_7$.

Also, $a_n>0$ and thus $a_n$ is bounded.

Is $0$ the highest lower bound of $a_n$ ? (Thinking)

evinda said:
First of all, it should be $d_{n+1}-d_n \geq 0$, right?

That is what we need for the sequence to be ascending.

evinda said:
This holds for $\left( 1+\frac{1}{n}\right)^{30} \geq 2$. Do we get from this that $1+\frac{1}{n} \geq 2^{\frac{1}{30}}$ although $30$ is even?

Then we would get that $n \leq \frac{1}{2^{\frac{1}{30}}-1}$.

So does this mean that the sequence hat its maximum value at the largest integer smaller than $\frac{1}{2^{\frac{1}{30}}-1}$ ?

How does it matter that $30$ is even?

It means that the maximum value is either at the larger integer smaller, or at the smallest integer bigger. (Thinking)

evinda said:
Since the sequence is descending for $n>7$, we have that $a_n< a_7$.

Also, $a_n>0$ and thus $a_n$ is bounded.

Don't we have that $a_7=a_8$? So $a_8 \not<a_7$ isn't it? (Worried)

evinda said:
Is $0$ the highest lower bound of $a_n$ ?

Yep. The sequence approaches 0 as $n\to\infty$, doesn't it? (Thinking)

Klaas van Aarsen said:
That is what we need for the sequence to be ascending.
How does it matter that $30$ is even?

It means that the maximum value is either at the larger integer smaller, or at the smallest integer bigger. (Thinking)

So the sequence is bounded below by $0$ and above by the smallest integer greater that $\frac{1}{2^{\frac{1}{30}}-1}$, right? (Thinking)
Klaas van Aarsen said:
Don't we have that $a_7=a_8$? So $a_8 \not<a_7$ isn't it? (Worried)
Yep. The sequence approaches 0 as $n\to\infty$, doesn't it? (Thinking)

Oh yes, right... (Nod)

evinda said:
So the sequence is bounded below by $0$ and above by the smallest integer greater that $\frac{1}{2^{\frac{1}{30}}-1}$, right?

Bounded below by $0$, yes. (Nod)

However, $\frac{1}{2^{\frac{1}{30}}-1}$ does not represent a value in the sequence does it?
Instead it shows at which $n$ the sequence changes from ascending to descending. (Worried)

Klaas van Aarsen said:
Bounded below by $0$, yes. (Nod)

However, $\frac{1}{2^{\frac{1}{30}}-1}$ does not represent a value in the sequence does it?
Instead it shows at which $n$ the sequence changes from ascending to descending. (Worried)

So the sequence hat its maximum value at the integer part of $\frac{1}{2^{\frac{1}{30}}-1}$, right? (Thinking)

evinda said:
So the sequence hat its maximum value at the integer part of $\frac{1}{2^{\frac{1}{30}}-1}$, right?

Either the integer part, or the integer part plus one. (Thinking)

## 1. How do I determine if a sequence is monotonic?

To check for monotonicity in a sequence, you need to observe the trend in the data points. If the values in the sequence are consistently increasing or decreasing, then the sequence is monotonic. If there are fluctuations or no clear trend, then the sequence is not monotonic.

## 2. What is the difference between a monotonically increasing and decreasing sequence?

A monotonically increasing sequence is one where the values consistently increase as you move along the sequence. On the other hand, a monotonically decreasing sequence is one where the values consistently decrease as you move along the sequence.

## 3. Can a sequence be both monotonically increasing and decreasing?

No, a sequence cannot be both monotonically increasing and decreasing. If the values in a sequence are constantly changing direction, then it is not monotonic.

## 4. How can I check for monotonicity in a large dataset?

One way to check for monotonicity in a large dataset is to plot the data points on a graph and visually inspect the trend. Another method is to calculate the difference between consecutive data points and check if it is consistently positive or negative.

## 5. What are some real-world applications of checking monotonicity in a sequence?

Checking monotonicity in a sequence is important in various fields such as economics, finance, and statistics. It is used to analyze trends in data, make predictions, and identify patterns. For example, in finance, checking the monotonicity of stock prices can help investors make informed decisions about buying or selling stocks.

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