Answer
$\{a_{n}\}$ diverges.
Work Step by Step
For odd n, $a_{n}=0$
For even n, $a_{n}=2\displaystyle \cdot\frac{n+1}{n}=2(1+\frac{1}{n})$
Let $\epsilon=1/2\gt 0.$
The sequence of even-indexed terms converges to $2$, and the sequence of odd-indexed terms converges to $0$.
If there were a limit L, then there would exist an index N, after which
$|L-a_{n}|\lt 1/2$
Every other (odd-indexed) term of the sequence is zero, so it follows that
$|L-0|\lt 1/2$
$L=(-\displaystyle \frac{1}{2},\frac{1}{2})$
But, since every other (even-indexed) term approaches 2, it would have to be that
$L\in(1.5,2.5)$ to satisfy the condition $|L-a_{n}|\lt\epsilon$.
L can't be in both these intervals, so it does not exist.
$\{a_{n}\}$ does not have a limit$\Rightarrow$it diverges.
