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Increasing/decreasing sequencesby cue928
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#1
May1711, 06:47 PM

P: 131

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. 


#2
May1711, 06:53 PM

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Yes, so the question is, which of the following two holds:
[tex]\frac{n}{e^n}\leq \frac{n+1}{e^{n+1}}~\text{or}~\frac{n}{e^n}\geq \frac{n+1}{e^{n+1}}[/tex] Try to simplify these expressions... 


#3
May1711, 07:09 PM

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


#4
May1711, 07:14 PM

P: 21

Increasing/decreasing sequences
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.



#5
May1711, 07:19 PM

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#6
May1711, 07:33 PM

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#7
May1711, 07:38 PM

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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 


#8
May1711, 07:43 PM

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You can start with this expression, [tex]\frac{n}{e^n}  \frac{n+1}{e^{n+1}}[/tex] and show that it is positive. That shows that a_{n} > a_{n+ 1}, or equivalently, that the sequence is decreasing. 


#9
May1711, 07:47 PM

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#10
May1711, 07:48 PM

P: 21




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