Determine whether the sequence with the given nth term is monotonic. Find the boundedness of the sequence. Otherwise, indicate DNE.
An = ne^(-n/2)
To determine whether a sequence is monotonic, we need to compare consecutive terms and check for an increasing or decreasing pattern. Let's begin by finding the general formula for the nth term (An) of the sequence.
Here, we are given the expression An = ne^(-n/2). This means that the nth term of the sequence can be calculated using this formula.
To analyze the monotonicity of the sequence, we need to check if the terms are strictly increasing or decreasing. For this, we can calculate the difference between consecutive terms to see if it is always positive (increasing) or negative (decreasing).
To find the difference between consecutive terms, we'll compare the (n+1)th term with the nth term by subtracting the nth term from the (n+1)th term:
An+1 - An = (n+1)e^(-(n+1)/2) - ne^(-n/2)
Now, simplify the expression:
An+1 - An = (n+1)e^(-(n+1)/2) - ne^(-n/2)
= e^(-(n+1)/2)((n+1) - n)
Simplifying further,
An+1 - An = e^(-(n+1)/2)
Since the difference between consecutive terms is solely determined by e^(-n/2), which is always positive, we can conclude that the terms of the sequence are strictly increasing. Thus, the sequence is monotonic and, specifically, monotonically increasing.
Now let's determine the boundedness of the sequence. A sequence is bounded if there exist upper and lower bounds, meaning there are finite values that the sequence does not exceed.
To check for boundedness, we can analyze the behavior of the terms as n approaches positive infinity.
As n gets larger, the term ne^(-n/2) approaches 0 because the exponential term e^(-n/2) becomes dominant. This indicates that the sequence is bounded above by 0.
Since the sequence is monotonically increasing and bounded above by 0, we can say that the sequence is bounded above but not bounded below.
In summary:
- The sequence defined by An = ne^(-n/2) is monotonically increasing.
- The sequence is bounded above by 0.
- The sequence is not bounded below.
To determine whether the sequence with the given nth term, An = ne^(-n/2), is monotonic, we can examine the behavior of successive terms of the sequence.
First, let's consider the difference between consecutive terms, An+1 and An:
An+1 - An = (n+1)e^(-(n+1)/2) - ne^(-n/2)
To analyze the monotonicity, we need to check whether the difference is always positive or always negative for all n.
Let's calculate this difference explicitly:
An+1 - An = (n+1)e^(-(n+1)/2) - ne^(-n/2)
= (n+1)e^(-n/2 - 1/2) - ne^(-n/2)
= (n+1)e^(-n/2 -1/2) - ne^(-n/2)
= (n+1)e^(-n/2 - 1/2) - ne^(-n/2 - 1/2)
Simplifying this expression, we obtain:
An+1 - An = e^(-n/2 - 1/2) ((n+1) - n)
= e^(-n/2 - 1/2)
The term e^(-n/2 - 1/2) is always positive for all n, so the sign of the difference An+1 - An is determined solely by the term (n+1) - n.
(n+1) - n = 1
Since 1 is positive, we can conclude that the difference An+1 - An is always positive for all n. Therefore, the sequence An = ne^(-n/2) is monotonically increasing.
Now, let's examine the boundedness of the sequence. To determine if the sequence is bounded, we need to find both an upper and lower bound.
Considering the nth term, An = ne^(-n/2), as n approaches infinity, the term will converge towards zero, since e^(-n/2) goes to 0 as n increases. Thus, the sequence is bounded above by zero.
Regarding the lower bound, we can observe that all terms in the sequence are positive, as both n and e^(-n/2) are positive for positive values of n. However, there is no specific lower bound since the terms continuously decrease towards zero but never reach it.
In summary:
- The sequence An = ne^(-n/2) is monotonically increasing.
- The sequence is bounded above by zero, but there is no specific lower bound.