Calculation of limsup and liminf Given the series $$\frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \cdots,$$
I don't know how the author gets 
$$\liminf_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \left(\frac{2}{3}\right)^n = 0$$
$$\liminf_{n\to\infty}a_n^{1/n} = \lim_{n\to\infty}\left(\frac{1}{3^n}\right)^{1/2n} = \frac{1}{\sqrt{3}}$$
$$\limsup_{n\to\infty}a_n^{1/n} = \lim_{n\to\infty}\left(\frac{1}{2^n}\right)^{1/2n} = \frac{1}{\sqrt{2}}$$
$$\limsup_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{1}{2}\left(\frac{3}{2}\right)^n = +\infty$$
And more generally, the book I'm using doesn't give a much precise definition on how to evaluate $\limsup$/$\liminf$.
 A: In general, if you have a sequence $\{b_1,b_2,\dots\}$, then
$$
\limsup b_i
$$
is defined to as follows: For each $i$, let $B_i=\{b_i,b_{i+1},\dots\}$. Then, 
$$
\limsup b_i=\lim_{i\rightarrow\infty}\left(\sup B_i\right).
$$
In other words, you take the sup's of each of the tails of the sequence and take the limit of those sup's.  In your case (for the first and last equalities), $b_i=\frac{a_{i+1}}{a_i}$.  $\liminf$ is defined similarly.
In your case (for the first and last equalities),
\begin{align}
b_1&=\frac{2}{3}&b_2&=\frac{3}{2^2}\\
b_3&=\frac{2^2}{3^2}&b_4&=\frac{3^2}{2^3}
\end{align}
so you can start to see the pattern:


*

*When the index on $b$ is odd, 
$$
b_{2i-1}=\frac{a_{2i}}{a_{2i-1}}=\frac{2^i}{3^i}=\left(\frac{2}{3}\right)^i.
$$

*When the index on $b$ is even,
$$
b_{2i}=\frac{a_{2i+1}}{a_{2i}}=\frac{3^i}{2^{i+1}}=\frac{1}{2}\left(\frac{3}{2}\right)^i.
$$
For a fixed $k$, the inf of the elements in $B_k$ is $0$ because $B_k$ contains higher and higher powers of $\frac{2}{3}$ (for all odd indices $i$ with $i>k$).  On the other hand, the supremum of $B_k$ because $B_k$ contains $\frac{1}{2}$ times higher and higher powers of $\frac{3}{2}$ (for even powers of $i$ with $i>k$).
I'm leaving the second and third cases as an exercise.
A: The series' general term is
$$a_n=\begin{cases}\cfrac1{2^{(n+1)/2}}\,,&\text{odd}\;n\\{}\\
\cfrac1{3^{n/2}}\,,&\text{even}\;n\end{cases}$$
In this case, we have the quotients:
$$\frac{a_{n+1}}{a_n}=\begin{cases}\cfrac{\frac1{2^{(n+1)/2}}}{\frac1{3^{n/2}}}=\cfrac1{\sqrt2}\left(\cfrac32\right)^{n/2}\,,\,&\text{even}\;n\\{}\\
\cfrac{\frac1{3^{n/2}}}{\frac1{2^{(n+1)/2}}}=\sqrt2\,\left(\cfrac23\right)^{n/2}\,,\,&\text{odd}\;n\end{cases}$$
and thus we get
$$\limsup_{n\to\infty}\frac{a_{n+1}}{a_n}=\sqrt2\,\cdot\,1=\sqrt2>1$$
and thus the series diverges. The same result you get with the $\;n\,-$ th root test.
