Question 953598
an = ne^-n


I believe you might mean:


An = n * e^-n


An meaning the nth term in the progression.


The fact that the exponent is -n, I would assume that the sequence is decreasing.


You can take a few values of n and show what is happening.


since it doesn't tell you that n has to be > 0, you should assume that n can be negative or positive or 0.


so, let's take some values of n between -3 and 3 and see what happens.


the following table shows what happens:


<pre>


n           n * e^(-n)         

-3          -3 * e^(-(-3)) = -3 * e^(3) = -60.25661...

-2          -2 * e^(-(-2)) = -2 * e^(2) = -14.77811...

-1          -1 * e^(-(-1)) = -1 * e^(1) = -2.71828...

0           0 * e^(0) = 0

1           1 * e^(-1) = .367879...

2           2 * e^(-2) = .270670...

3           3 * e^(-3) = .149361...

100         100 * e^(-100) = 3.720075... * 10^(-42)

</pre>


it's clear that the function gets very small the further negative you go with the value of n and that the function stays positive and approaches 0 the further positive you get.


it does not appear that there is a vertical asymptote.


in order for there to be a vertical asymptote, the denominator of a rational equation needs to be equal to 0 at some point.


n * e^(-n) is the same as n / e^n


the denominator is e^n, but the denominator can never be equal to 0 because e^n can never be equal to 0.


when n = 0, e^n is equal to e^0 which is equal to 1.


so, there is no vertical asymptote, even though the function gets extremely negative as n gets more negative.


in fact, it gets so negative that you won't be able to see it on the graph after n becomes more negative than -10 or so unless you scale into the millions.


for example:


when n = -1000, n * e^(-n) becomes -1000 * e^1000 which becomes so large a negative value that the calculator can't handle it.


when n = -100, n * e^(-n) becomes -100 * e^100 which becomes -2.688117... * 10^45.


that's a very large negative number, but the function does not approach infinity as n approaches a certain number, so there is no vertical asymptote.


is there a horizontal asymptote?


that can be found by increasing n as large as it can get.


as n approaches infinity, the function approaches 0.


you get n * e^-n becomes n / e^n.


as n approached infinity, the function gets smaller and smaller.


you can see this clearer by dividing numerator and denominator by n to get:


(n/n) / ((e^n)/n) which results in 1 / ((e^n)/n)


(e^n)/n  will get larger and larger as n gets larger.


the numerator stays the same at 1.


the function gets smaller and smaller and approaches zero as n grows larger and larger and approaches infinity.


for example, when n = 1000, n/e^n is equal to 1000 / e^1000 which is equal to 0.


it's not really equal to 0, but the calculator can't handle a number that small so the calculator shows 0.


when n = 100, n/e^n is equal to 100 / e^100 which is equal to 3.72007... * 10^(-42).


that's a very small number that's pretty close to 0.  


as n gets larger, it gets close to 0 but never reaches 0.


so you have a function that approaches minus infinity as n gets more negative and you have a function that approaches 0 as n gets more positive.


a graph of the equation is shown below:


in the graph, n is replaced by x in order to satisfy the requirements of the graphing software.


otherwise the equation is the same and when i talk about n, i'm also talking about x because they represent the same thing.


there are 2 graphs to show you how the function behaves.


the first graph is a view from a distance to see what happens on the negative side of the graph as n gets more negative.


the second graph is a close up view to see what happens on the positive side of the graph as n gets more positive.


don't forget that x represent n in the graph.


<img src = "http://theo.x10hosting.com/2015/030701.jpg" </>


<img src = "http://theo.x10hosting.com/2015/030702.jpg" </>


so what is the answer to your question?


the question was:


Decreasing or increasing ? an=ne^-n
I need not only answer but explanation too.


the function increases to a maximum point and then decreases thereafter as it approaches the horizontal asymptote of y = 0.


how do you explain that?


good question.


i would say you can say that n * e^-n is the same as n / e^n and it is clear clear that e^n rises dramatically faster than n as the value of n gets larger.


as n approaches infinity, the function n / e^n therefore approaches 0.   


the function is decreasing after it reaches it's maximum point at somewhere between n = 0 and n = 2.


in fact, it appears that the maximum value is as n = 1.


when n = 1, the function is equal .3678794412...


when n = .999, the function is equal to .3678792571


when n = 1.001, the function is equal to .3678792574...


I can't swear to it, but n = 1 does appear to be the maximum point of the function.


so the function reaches a maximum at n = 1 and then decreases thereafter and approaches 0 but never quite touches it.


the short answer.


the function increases for a while and then decreases.