How to evaluate $ \lim_{{k\to\infty} {p\to\infty}}\frac{a_{k,p}}{b_{k,p}}$? 
Define
  $$
\begin{align}
a_{k,p}=&\sum_{n=1}^{k} (2^{n}-1)\\
+&\sum_{n=1}^{k}{(2^{2n}-2^n-1)}\\
+&\sum_{n=1}^{k}{(2^{3n}-2^{2n}-2^n-1)}\\
+&\sum_{n=1}^{k}{(2^{4n}-2^{3n}-2^{2n}-2^n-1)}\\
+&...\\
+&\sum_{n=1}^{k}{(2^{pn}-2^{n(p-1)}-2^{n(p-2)}-...-2^{3n}-2^{2n}-2^n-1)}
\end{align}
$$
  and
  $$
b_{k,p}={\frac14\times{[2^{pk}-2^{k(p-1)}-2^{k(p-2)}-...-2^{3k}-2^{2k}-2^k}]^2}
$$
How to evaluate
  $$
\lim_{{k\to\infty}
{p\to\infty}}\frac{a_{k,p}}{b_{k,p}}?$$

 A: ---- Direct Proof Using Squeeze Theorem ----
Using the following simplifications for $a_{k,p}$ and $b_{k,p}$:
$$ a_{k,p}=\sum_{n=1}^k\left[2^{pn}-\sum_{m=1}^{p-2}m2^{(p-1-m)n}-p\right]\!;\ and\ \ b_{k,p}=\frac{1}{4}\left(2^{pk}-\sum_{n=1}^{p-1}2^{nk}\right)^2, $$
one can use the geometric series and basic inequality results to obtain:
$$ \frac{a_{k,p}}{b_{k,p}}=\frac{\displaystyle{\sum_{n=1}^k\left[2^{pn}-\sum_{m=1}^{p-2}m2^{(p-1-m)n}-p\right]}}{\displaystyle{\frac{1}{4}\left(2^{pk}-\sum_{n=1}^{p-1}2^{nk}\right)^2}}\leq\frac{\displaystyle{4\sum_{n=1}^k2^{pn}}}{\displaystyle{\left(2^{pk}-2^k\frac{2^{k(p-1)}-1}{2^k-1}\right)^2}} $$
$$ \leq\frac{4\cdot\displaystyle{2^p\frac{2^{pk}-1}{2^p-1}}}{\displaystyle{\left(2^{pk}-2^k\frac{2^{k(p-1)}}{2^k-1}\right)^2}}\leq\frac{\displaystyle{4\cdot2^{pk}\frac{1}{1-2^{-p}}}}{\displaystyle{\left(2^{pk}\left(\frac{2^k-2}{2^k-1}\right)\right)^2}}\leq\frac{8\cdot2^{pk}}{\displaystyle{\left(2^{pk}\left(\frac{1}{2}\right)\right)^2}}=\frac{32}{\displaystyle{2^{pk}}}\equiv x_{k,p}. $$
It can easily be shown that $\lim(x_{k,p})=0$. Because $0<\frac{a_{k,p}}{b_{k,p}}\leq x_{k,p}$ for all positive $k,p$, it follows from the Squeeze Theorem that the limit in question is 0.
---- Originally Proposed Route Using Ratio Test ----
This route to arriving at a proof is still workable but requires a lot more effort and scratch paper; it can indeed be shown that $L<1$.
Perhaps this question can be transformed into a different one using the Ratio Test.  That is, define a new sequence $X$ as the ratio:
$$X=(x_{k,p})\equiv\left(\frac{a_{k+1,p+1}\cdot b_{k,p}}{b_{k+1,p+1}\cdot a_{k,p}}\right).$$
If one can show that $\lim(x_{k,p})=L$ exists and that $0<L<1$, then $\lim\left(\frac{a_{k,p}}{b_{k,p}}\right)=0$.
A: Since $a_{k,p}$ grows (for large $p$ and $k$) like $2^{k(p-1)}$ and $b_{k,p}$ grows like 
$\frac14 2^{2kp}$ the ratio $\frac{a_{k,p}}{b_{k,p}}$ goes to zero.  
Are you sure you meant to have the square in the definition of $b_{k,p}$?
