Answer:
if the sequence is:
12, 13, 13, 14, 14 etc, and each term keeps growing up, the sequence obviusly diverges.
Now, if the sequence is
1/2, 1/3, 1/3, 1/4, 1/4, 1/5 , 1/5
so the terms after the first one repeat, we could group the terms with the same denominator and get:
1/2, 2/3, 2/4, 2/5..... etc.
So the terms after the first one are aₙ = 2/n.
Now, a criteria to see if a sequence converges if seing if:

and here we have;

that obviusly tends to zero, so we can conclude that this sequence converges.
then the limit is:
There exist a n' such that for any n > n' then IL -aₙI < ε
where L is the limit
I2/n - 0I = I2/nI < ε
then this is true if n > 2/ε = n'