MHB How to check monotony of sequence

[evinda]

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)

 

[I like Serena]

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)

 

[evinda]

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)

 

[evinda]

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)

 

[I like Serena]

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

Yep. (Nod)

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)

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)

 

[HallsofIvy]

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

 

[evinda]

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)

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)

 

[I like Serena]

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.

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)

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)

Is $0$ the highest lower bound of $a_n$ ?

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

 

[evinda]

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)

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)

 

[I like Serena]

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)

 

[evinda]

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)

 

[I like Serena]

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)