Max Value of a_n: Find n for Maximum Value

[Saitama]

Homework Statement

(These may not be the exact wordings as it was asked by my friend and I do not have the exact question)
Find the value of n for which a_n is maximum where
$$a_n=\frac{1000^n}{n!}$$

(Ans: n=999)

Homework Equations

The Attempt at a Solution

I don't think using calculus in this type of question would help so I have put it in Precalc section. I am clueless here, how should I begin with this.

 

[CompuChip]

You could try looking at $$\frac{a_{n + 1}}{a_n}$$

 

[rollingstein]

I took log of both sides and then used the Stirling Approximation.

Then differentiate with respect to n and set that to zero.

I get n=1000 (but there must be some ugliness because of the discontinuity etc.)

 

[CompuChip]

If you follow my hint, you will find a cleaner approach.

 

[Saitama]

You could try looking at $$\frac{a_{n + 1}}{a_n}$$

This is equal to $\frac{1000}{n+1}$.
But I am still clueless.

 

[rollingstein]

This is equal to $\frac{1000}{n+1}$.
But I am still clueless.

If the series in increasing what's the value of the ratio. If it is decreasing what's the value of the ratio?

@CompuChip: Yes, that's cleaner.

 

[rollingstein]

In any case shouldn't 999 and 1000 both be acceptable answers?

 

[Saitama]

If the series in increasing what's the value of the ratio. If it is decreasing what's the value of the ratio?

If the series increases,its greater than 1 and less than 1 if it decreases.

 

[rollingstein]

If the series increases,its greater than 1 and less than 1 if it decreases.

You still don't see the answer?

 

[Saitama]

You still don't see the answer?

Honestly, no.

 

[CompuChip]

If the series increases,its greater than 1 and less than 1 if it decreases.

So what happens at the maximum?

 

[Saitama]

Can you tell me to which area of mathematics these type of questions belong?

 

[Saitama]

@CompuChip: Sorry if this is annoying you but I really have no idea.

 

[CompuChip]

Look at the bit that I quoted: you say that if the series is increasing, then the ratio between consecutive elements is > 1. If it is decreasing, then the ratio is < 1.

Now imagine a series with a maximum. Draw one on a piece of paper, if it helps. What can you say about the increasing / decreasing on either side of the maximum? What does that tell you about the ratio "at" the maximum?

 

[Saitama]

Before the maximum, series increases. After that it decreases.

 

[CompuChip]

So before the maximum $a_{n+1}/a_n > 1$ and after that $a_{n+1}/a_n < 1$. What happens in between?

 

[SammyS]

In any case shouldn't 999 and 1000 both be acceptable answers?

Well it is pretty easy to show that a₉₉₉ = a₁₀₀₀ .

 

[Opus_723]

But wait!

$\frac{1000^{999.5}}{999.5!}$

is larger than either of those...

And now we go to scary places.

 

[Saitama]

So before the maximum $a_{n+1}/a_n > 1$ and after that $a_{n+1}/a_n < 1$. What happens in between?

Equal to 1. Therefore n=999.

I plugged a_n in wolframalpha with n=999 and n=1000. The results are same but how can I show that?

But wait!

$\frac{1000^{999.5}}{999.5!}$

is larger than either of those...

And now we go to scary places.

As I do not have the exact question, it should be mentioned that n is an integer.

 

[Dick]

Equal to 1. Therefore n=999.

I plugged a_n in wolframalpha with n=999 and n=1000. The results are same but how can I show that?

n=999 gives 1000^999/999!. n=1000 gives 1000^1000/1000!. 1000!=999!*1000.

 

[Saitama]

n=999 gives 1000^999/999!. n=1000 gives 1000^1000/1000!. 1000!=999!*1000.

Thanks Dick!

 

[Saitama]

You could try looking at $$\frac{a_{n + 1}}{a_n}$$

I still want to ask one question. How you came up with this or how did this pop up in your mind? I am interested to know how you think about these type of problems.

 

[ehild]

Having a sequence of positive numbers, if nth is the greatest among them, it must be greater then its neighbours, so an+1≤an and an-1≤an

ehild

 

[Ray Vickson]

Equal to 1. Therefore n=999.

I plugged a_n in wolframalpha with n=999 and n=1000. The results are same but how can I show that?



As I do not have the exact question, it should be mentioned that n is an integer.

You already have all you need: you have a_1000/a_999 = 1, so a_999 = a_1000 !

 

[CompuChip]

You had already shown that
$$\frac{a_{n+1}}{a_n} = \frac{1000}{n + 1}$$
If you set that equal to 1, you can solve for n quite straightforwardly.

I guess this is one of those things where a bit of experience helps... I was initially looking at the limit for $n \to \infty$ and thought of the ratio test. Then I noticed that in the ratio the factorial disappeared, and somehow made the connection to your actual question.

 

[Saitama]

You had already shown that
$$\frac{a_{n+1}}{a_n} = \frac{1000}{n + 1}$$
If you set that equal to 1, you can solve for n quite straightforwardly.

I guess this is one of those things where a bit of experience helps... I was initially looking at the limit for $n \to \infty$ and thought of the ratio test. Then I noticed that in the ratio the factorial disappeared, and somehow made the connection to your actual question.

Thanks for the insight CompuChip!

 

[ehild]

If a_n is the greatest, then

$\frac{a_{n+1}}{an}\leq 1$

and

$\frac{a_{n-1}}{an}\leq 1$ .$\frac{n}{1000}\leq 1$, $\frac{1000}{n+1}\leq 1$,

that is

$999 \leq n \leq 1000$.

The 999-th term is he same as the 1000th, as Dick pointed out. There are two solutions, both 999 and 1000.

ehild

 

[Saitama]

If a_n is the greatest, then

$\frac{a_{n+1}}{an}\leq 1$

and

$\frac{a_{n-1}}{an}\leq 1$ .


$\frac{n}{1000}\leq 1$, $\frac{1000}{n+1}\leq 1$,

that is

$999 \leq n \leq 1000$.

The 999-th term is he same as the 1000th, as Dick pointed out. There are two solutions, both 999 and 1000.

ehild

Thanks ehild!

Posts like this always help when I look back at the old threads I have posted. :)

 

[Dick]

The 999-th term is he same as the 1000th, as Dick pointed out.
ehild

In all fairness, SammyS pointed that out. I just chirped in with the reason why it was true. I was late to the game.

 

[ehild]

Ah, yes, it was SammyS and also rollingstein...

ehild