Whether the sequence is increasing, decreasing, not monotonic and bounded or not.

The sequence is not monotonic and not bounded.

Definition used:

(1) The sequence {an} is increasing if an<an+1 for all n≥1 . That is, a1<a2<a3<... .

(2) The sequence {an} is decreasing and an>an+1 for all n≥1 . That is, a1>a2>a3>... .

(3) If the sequence is either increasing or decreasing, then the sequence is called monotonic; otherwise it is not monotonic.

(4) If {an} is the sequence with m≤an≤M for m,M∈ℕ , then the sequence is bounded.

Theorem used:

If limn→∞|an|=0 , then limn→∞an=0 . (1)

Given:

The sequence is an=n(−1)n (2)

Calculation:

Check whether the sequence is increasing, decreasing or not monotonic.

The first five terms of the sequence is as follows:

{n(−1)n}={1(−1)1,2(−1)2,3(−1)3,4(−1)4,5(−1)5...}   ={1(−1),2(1),3(−1),4(1),5(−1)...}   ={−1,2,−3,4,−5,...}

Since a1<a2 but a2>a3 , the sequence is neither increasing nor decreasing.

Therefore, by definition (3), the sequence is not monotonic.

Check whether the sequence is bounded or not.

Obtain the limit of the sequence (the value of the term an as n tends to infinity).

limn→∞an=limn→∞n(−1)n

The absolute value of n(−1)n is,

|n(−1)n|={|n(1)|if n is even|n(−1)|if n is odd=n

The absolute value of n(−1)n when n→∞ is obtained as follows:

limn→∞|n(−1)n|=limn→∞(n)=∞

Since limn→∞|an|=∞ and by using the theorem (1), the sequence {an} is divergent.

That is, the sequence is not bounded.

Therefore, it can be concluded that the sequence is not monotonic and not bounded.