Calculus of a Single Variable, Chapter 9, 9.1, Section 9.1, Problem 55

a_n=n e^(-n/2)
Monotonicity
First five terms of the sequence are
a_1=e^(-1/2)=0.6065
a_2=2e^-1=0.7358
a_3=3e^(-3/2)=0.6693
a_4=4e^-2=0.5413
a_5=5e^(-5/2)=0.4104
We can see that after the second term, the terms are decreasing so it is possible that the whole sequence is monotonically decreasing. Let us verify that.
a_n>a_(n+1)
n e^(-n/2)>(n+1)e^(-(n+1)/2)=(n+1)e^(-n/2)e^(-1/2)
Divide the inequality by e^(-n/2). We can do that because e^(-n/2)>0, forall n in NN.
n>(n+1)e^(-1/2)=n e^(-1/2)+e^(-1/2)
Divide by n. We can do that because n>0.
1>e^(-1/2)+e^(-1/2)/n
Since e^(-1/2)<1, we can find sufficiently large n such that the above inequality holds. In this case n=2.
1>e^(-1/2)+e^(-1/2)/2=0.9098
Therefore, forall n geq 2 (a_n>a_(n+1)) which means that the sequence is monotonically decreasing.
Boundedness
We have shown that the sequence is monotonically decreasing from second term onwards. This means that the second terms is also maximum of the sequence. In other words the sequence is bounded from above by a_2=2e^-1.
On the other hand if we look at the sequence a_n=n e^(-n/2), we see that all of its terms are positive. This is because n>0 and exponential function is always positive so e^(-n/2)>0 and the product of two positive numbers is itself positive.
Therefore, a_n>0, forall n in NN i.e. the sequence is bounded by zero from below.
We can conclude that forall n in NN, a_n in [2e^(-1),0) i.e. the sequence is bounded from both below and above.
The image below shows first 20 terms of the sequence. Both boundedness and monotonicity can clearly be seen on the image.