Finding lim sup I am unable to find a way to solve this. I have tried using definition of lim sup but got no where.

Consider the series: $\sum^\infty_{n=1} a_n = \frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\cdots$
Then $\limsup_{n\to\infty} \frac{a_{n+1}}{a_n}=$

Edit: This is the work I have done yesterday night. My apologies for not showing my work.
I have looked at the sequence $\frac{a_{n+1}}{a_{n}}$. I observed that it looks like $\frac{2}{3}, \frac{3}{2^2}, \frac{2^2}{3^2},\frac{3^2}{2^3},\cdots$.
So, the general terms look like $(\frac{2}{3})^n$ for $n$ odd and $\frac{1}{2}(\frac{3}{2})^n$ for $n$ even.
I looked at the supremums of the tail sequences and then decided to find their infimum. The sequence as I found has $\infty$ as the supremum of each tail sequence since the sequence for $n$ even is monotonically increasing. Thus I inferred that the limsup should be $\infty$. However, the answer was given to be $\frac{1}{2}$.
This sequence I found is of course a mistake as the exponents I have taken are wrong.
The sequence actually looks like $\frac{a_{n+1}}{a_n}=\cases{(\frac{2}{3})^{\frac{n+1}{2}}& n is odd\\ \frac{1}{2}(\frac{3}{2})^\frac{n}{2}& n is even}$
But still, the sequence of supremum of the tails consists only of $\infty$ and hence the infimum of the supremums is $\infty$. What am I missing?

Your argument is perfectly fine and the "bandit" in this one is the author of the problem for making it unclear on how $a_{n}$ is to be expressed. Something I have found true for some authors is that; when in doubt a Mathematics textbook is always written in "three's".

If  instead of treating $a_{n}$ to be a sequence of alternating reciprocal powers of $2$ and $3$, we take $a_{n}$ to be a sequence of equal powers, as in $a_{n}=\frac{1}{2^{n}}+\frac{1}{3^{n}}$, then we get an alternate (and in hindsight a true) description of the series in question. By the following sequence of calculations we reach $\lim \sup \frac{a_{n+1}}{a_{n}}=\frac{1}{2}$:
First by rewriting $a_{n}$ on a common denominator we get
$a_{n}=\frac{3^{n}+2^{n}}{6^{n}}$ and so
$$\frac{a_{n+1}}{a_{n}}=\frac{2^{n+1}+3^{n+1}}{6(2^{n}+3^{n})}.$$
Dividing numerator and denominator by $3^{n+1}$ we arrive at
$$\frac{a_{n+1}}{a_{n}}=\frac{(\frac{2}{3})^{n+1}+1}{2((\frac{2}{3})^{n}+1)}.$$
Now since
$$(\frac{2}{3})^{n},(\frac{2}{3})^{n+1}\to 0 \ \text{as} \ n \to \infty$$
we get that
$$\lim \frac{a_{n+1}}{a_{n}}=\lim \frac{(\frac{2}{3})^{n+1}+1}{2((\frac{2}{3})^{n}+1)}=\frac{1}{2}.$$
Because $\lim \sup x_{n} = \lim x_{n}$ for a sequence $x_{n}$ which converges, we can infer that
$$\lim_{n\to\infty} \sup \frac{a_{n+1}}{a_{n}}=\frac{1}{2}$$.
