**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.