Increasing/decreasing sequences

[cue928]

I have been asked to find if the following sequence is increasing or decreasing:
an = ne^-n

So I first multiplied thru by e^n to get: n/e^n
Then, I did n+1, so (n+1)*e^-(n+1). I moved the negative exponent to the bottom to get
(n+1)/(e^(n+1))

I guess my first question is did I start off correctly.

 

[micromass]

Yes, so the question is, which of the following two holds:

$$\frac{n}{e^n}\leq \frac{n+1}{e^{n+1}}~\text{or}~\frac{n}{e^n}\geq \frac{n+1}{e^{n+1}}$$

Try to simplify these expressions...

 

[Mark44]

I have been asked to find if the following sequence is increasing or decreasing:
an = ne^-n

So I first multiplied thru by e^n to get: n/e^n

What you have is equal to ne^-n, but you didn't get it by multiplying by eⁿ, which would have been wrong to do.

e^-n is equal to 1/eⁿ. What you did was rewrite the original expression in different form, using a property of exponents.

Then, I did n+1, so (n+1)*e^-(n+1). I moved the negative exponent to the bottom to get
(n+1)/(e^(n+1))

?
You're working with an expression, ne^-n. Unlike an equation, there are only a few things you can do with an equation. The only number you can add to an expression is 0 - otherwise you will change the value of the expression. The only number you can multiply by is 1 - which can take many forms, but it's still 1. If you multiply by something other than 1, you will get a new expression that is not equal to the one you started with.

So in particular, you can't just multiply by n + 1, and you can't just decide to change the exponent from -n to -(n + 1). Those are not valid operations here.

I guess my first question is did I start off correctly.

 

[Maxter]

I don't know if you are allowed to do this, but the standard way is to take the derivative of f(x)=xe^-x and show that it's negative for x>1, thus is a decreasing function on that interval. Since a_n=f(n), this gives that the sequence is decreasing.

 

[Mark44]

I don't know if you are allowed to do this, but the standard way is to take the derivative of f(x)=xe^-x and show that it's negative for x>1, thus is a decreasing function on that interval. Since a_n=f(n), this gives that the sequence is decreasing.

This is one way. micromass is showing another way that more directly shows that the sequence is decreasing.

 

[Maxter]

This is one way. micromass is showing another way that more directly shows that the sequence is decreasing.

Right, which means you have to start with one side, and work to the other. It's not clear how to do that directly. Technically, you cannot start with either of those assumptions and simplify, though doing that might give you a hint with how to work in a more direct manner.

 

[micromass]

Right, which means you have to start with one side, and work to the other. It's not clear how to do that directly. Technically, you cannot start with either of those assumptions and simplify, though doing that might give you a hint with how to work in a more direct manner.

In this particular case, my approach is not that hard. Just bring the n+1 and the eⁿ to the other side, and you'll see something nice popping up!

I do agree that your approach would be best in almost every other case, as derivatives are quite easy to handle In fact, I would have mentioned the derivative thingy, but I forgot about it

 

[Mark44]

Right, which means you have to start with one side, and work to the other.

Not necessarily.
You can start with this expression,
$$\frac{n}{e^n} - \frac{n+1}{e^{n+1}}$$
and show that it is positive. That shows that a_n > a_{n+ 1}, or equivalently, that the sequence is decreasing.

It's not clear how to do that directly. Technically, you cannot start with either of those assumptions and simplify, though doing that might give you a hint with how to work in a more direct manner.

 

[Maxter]

Not necessarily.
You can start with this expression,
$$\frac{n}{e^n} - \frac{n+1}{e^{n+1}}$$
and show that it is positive. That shows that a_n > a_{n+ 1}, or equivalently, that the sequence is decreasing.

RIGHT, forgot about doing that. Thanks!

 

[Maxter]

In this particular case, my approach is not that hard. Just bring the n+1 and the eⁿ to the other side, and you'll see something nice popping up!

I do agree that your approach would be best in almost every other case, as derivatives are quite easy to handle In fact, I would have mentioned the derivative thingy, but I forgot about it

And I forgot about the subtraction technique!