Answer
Proof given below.
Work Step by Step
Let $\epsilon \gt 0.$
Since $a_{2k}\rightarrow L,$ there exists an index $N_{1}$ after which
$(2k \gt N_{1})\Rightarrow |a_{2k}-L| \lt \epsilon$
Since $a_{2k+1}\rightarrow L,$ there exists an index $N_{2}$ after which
$(2k+1 \gt N_{2})\Rightarrow |a_{2k+1}-L| \lt \epsilon$
We find $N$ to be the greater of the two, $N=\displaystyle \max\{N_{1},N_{2}\}$, for which,
since n is either odd or even, $n=\left\{\begin{array}{ll}
2k & or\\
2k+1 &
\end{array}\right.$ ,
it follows that
$(n \gt N)\Rightarrow |a_{n}-L| \lt \epsilon$
In other words (by the definition), $a_{n}\rightarrow L$
