TITLE: Compute limsup liminf
QUESTION [1 upvotes]: Possible Duplicate:
Compute $\liminf (a_k)^{1/k}$ $\limsup (a_k)^{1/k}$ $\liminf (a_{k+1}/a_k)$ and $\limsup (a_{k+1}/a_k)$ as $k \rightarrow \infty$ 

For the series 
$$\frac{1}{2} + 1 + \frac{1}{8} + \frac{1}{4} + \frac{1}{32} + \frac{1}{16} + \frac{1}{128} + \frac{1}{64} +...$$ 
Compute 

$\liminf_{k\to\infty}(a_k)^{1/k}$
$\limsup_{k\to\infty}(a_k)^{1/k}$
$\liminf_{k\to\infty}(a_{k+1}/a_k)$ and 
$\limsup_{k\to\infty}(a_{k+1}/a_k)$.

This is geometric series. But this does not order. So how I can find?

REPLY [3 votes]: The ratios $\dfrac{a_{k+1}}{a_k}$ are easiest to deal with. Note that they are alternately $2$ and $\dfrac{1}{8}$. Now the $\liminf$ and $\limsup$ should be easy to write down. 
For the roots, call the terms of our sequence $a_1,a_2,a_3,a_4,\dots$. 
Compute $a_k^{1/k}$ for the first few terms. It will tell you what's going on. 
But we are a little impatient, so will go to formulas. Note that our sequence goes 
$$2^{-1}, 2^{-0}, 2^{-3}, 2^{-2}, 2^{-5}, 2^{-4}, 2^{-7}, 2^{-6},\dots.$$  
So if $k$ is odd, then $a_k=2^{-k}$. If $k$ is even, then $a_k=2^{-(k-2)}=\dfrac{2^{-k}}{4}$.
Now take $k$-th roots. If $k$ is odd, then $a_k^{1/k}=2^{-1}$.
If $k$ is even, then $a_k^{1/k}=\dfrac{2^{-1}}{4^{1/k}}$. Now it should not be hard to answer the questions about $\liminf$ and $\limsup$.
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