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 $$ \lim_{{k\to\infty} {p\to\infty}}\frac{a_{k,p}}{b_{k,p}}?$$

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    $\begingroup$ One can introduce some notations to make the limit more readable. $\endgroup$ – Jack Jul 21 '17 at 21:43
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    $\begingroup$ And you certainly want to add some parentheses properly. $\endgroup$ – Jack Jul 21 '17 at 21:44
  • $\begingroup$ Dear @Jack can you edit for me? $\endgroup$ – MathLover Jul 21 '17 at 21:44
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    $\begingroup$ I was trying to edit it but I'm not sure I understand the expression correctly. First of all, do you mean $(\sum_{n=1}^k2^n)-1$ or $\sum_{n=1}^k(2^n-1)$ in the numerator? $\endgroup$ – Jack Jul 21 '17 at 21:46
  • $\begingroup$ @Jack second. For all. $\endgroup$ – MathLover Jul 21 '17 at 21:47

---- 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 $\lim\left(\frac{a_{k,p}}{b_{k,p}}\right)=0$.

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  • $\begingroup$ I think, in my limit It must be $L>1$, definetly $\endgroup$ – MathLover Jul 27 '17 at 11:15
  • $\begingroup$ See above for an alternative route to obtaining your answer. However, if the Ratio Test is used, it will indeed be found that $L<1$ (not $L>1$). This requires a fair amount of scratch paper and use of the geometric series to find out however. $\endgroup$ – Stephen K. Jul 27 '17 at 18:59
  • $\begingroup$ Hmm...You worked hard. I'm reviewing the solution.. $\endgroup$ – MathLover Jul 27 '17 at 20:15

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 $b_{k,p}$?

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  • $\begingroup$ Yes I am sure. Yes I know finite large number $p$ and $k$ limit equal $0$. but here $p,k-> \infty$ $\endgroup$ – MathLover Jul 21 '17 at 22:45
  • $\begingroup$ In that case the limit is in fact zero. In general, if for fixed $p$ above some $p_0$ the limit of $f(p,k)$ as $k\to\infty$ is $L$, and if for fixed $k$ above some $k_0$ the limit of $f(p,k)$ as $p\to\infty$ is $L$, then the limit as they both go to infinity is$L$. Here, $L$ is zero. $\endgroup$ – Mark Fischler Jul 25 '17 at 18:37
  • $\begingroup$ It's ok.Finally, Is general limit equal to $1$? $\endgroup$ – MathLover Jul 27 '17 at 11:13

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