Whether the sequence converges or diverges and obtain the limit if the sequence converges.

The sequence converges to the limit 0.

Given:

The sequence is {11,13,12,14,13,15,14,16,...} .

Here, a1=11,a2=13,a3=12,a4=14,a5=13,a6=15,a7=14,a8=16,... .

Definition used:

If an is a sequence and limn→∞an exists, then the sequence an is said to be converges; otherwise it diverges.

Calculation:

Consider the sequence an={11,13,12,14,13,15,14,16,...} .

Divide the sequence an into two subsequences which are {a1,a3,a5,a7...} and

{a2,a4,a6,a8...} .

That is, the odd terms of the sequence is a2n−1={11,12,13,14,...} and the even terms of the

sequence is a2n={13,14,15,16,...} .

Consider the subsequence a2n−1={11,12,13,14,...} .

Here, it is observed that the denominator of every term is the number of the term n.

Therefore, the formula for general term a2n−1 of the sequence is 1n .

Obtain the value of the term a2n−1 when n tends to infinity. That is,

limn→∞a2n−1=limn→∞(1n)             =1∞             =0

Since the subsequence limn→∞a2n−1 exists and by using the definition, it can be concluded that the subsequence converges to the limit 0.

Consider the subsequence a2n={13,14,15,16,...} .

That is, a2n={12+1,12+2,12+3,12+4,...} .

Therefore, the formula for general term a2n−1 of the sequence is 12+n .

Obtain the value of the term a2n when n tends to infinity.

limn→∞a2n=limn→∞(12+n)           =limn→∞(1)limn→∞(2+n)           =1∞           =0

Since the subsequence limn→∞a2n exists and by using the definition, it can be concluded that  the subsequence converges to the limit 0.

Hence, the both the subsequences a2n−1 and a2n converges to the limit 0. Thus, the original sequence an converges to the same limit zero.

Therefore, the sequence converges to the limit 0.