What is a monotonic sequence and how do you determine its boundedness?

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Homework Statement


Determine whether the sequence with the given nth term is monotonic. Find the boundedness of the sequence.
[tex] a_n = ne^{-n/2} [/tex]

Homework Equations


I don't know


The Attempt at a Solution


I have absolutely no idea what a monotonic sequence is or how to find the boundedness of a sequence. I've tried researching it but I'm still confused. Any help would be greatly appreciated.
 
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Hi Barbados_Slim! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)

"monotonic" means that it only goes one way …

either it never decreases, or it never increases …

see http://en.wikipedia.org/wiki/Monotonic" :wink:

I don't know what "boundedness" means … it seems rather vague. :redface:
 
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Well thank you for your prompt answer. I hope you don't mind but I have another question.
[tex] \sum_{k=1}^{\infty} \frac {1} {k(k+1)}[/tex]
is an example of a telescoping series. Find a a formula for the general term [itex]S_n[/itex] of the sequence of partial sums.
I've reached the conclusion that the formula for the general term is
[tex] \frac {k} {k+1}[/tex]
but webassign is telling me that it is the wrong answer. Can anyone help, it would be grealty appreciated.
 
Expand using partial fractions: [tex]\frac{1}{k(k+1)}=\frac{\,A\,}{k}+\frac{B}{k+1}[/tex]

Find A & B.
 
By boundedness, are there certain values which the values of the sequence never get larger (an upper bound) or smaller (a lower bound) than?
 
I figured out the problem with the telescoping series. I was just using the wrong letter, I used "k" instead of "n". As for the other problem about the boundedness. I believe that boundedness refers to certain values that the sequence never gets larger or smaller than, like jhae2.718 said. The graph of the function doesn't appear to be bounded but I got the wrong answer when I said that the bounds do not exist. I think the answer might be zero because
[tex] \lim_{n \rightarrow \infty} ne^{-n/2} = 0[/tex]
Thank you so much for your help.