Limit of quotient of summations involving special binomial coefficients

Question

Find the limit, when $n$ tends to infinity, of $$ \frac{\displaystyle\sum_{k=0}^n\binom{2n}{2k}3^k} {\displaystyle\sum_{k=0}^{n-1}\binom{2n}{2k+1}3^k} $$

Please Help Me to solve the problem given and i have no idea to do since i have done such this one and need help also many hint of course.

Answer 1

Note that $$ \sum_{k=0}^n\binom{2n}{2k}3^k-\sqrt3\sum_{k=0}^{n-1}\binom{2n}{2k+1}3^k=(\sqrt3-1)^{2n} $$ and $$ \sum_{k=0}^n\binom{2n}{2k}3^k+\sqrt3\sum_{k=0}^{n-1}\binom{2n}{2k+1}3^k=(\sqrt3+1)^{2n} $$ Therefore, $$ \begin{align} \frac{\displaystyle\sum_{k=0}^n\binom{2n}{2k}3^k}{\displaystyle\sum_{k=0}^{n-1}\binom{2n}{2k+1}3^k} &=\sqrt3\,\frac{(\sqrt3+1)^{2n}+(\sqrt3-1)^{2n}}{(\sqrt3+1)^{2n}-(\sqrt3-1)^{2n}}\\ &=\sqrt3\,\frac{1+\left(\frac{\sqrt3-1}{\sqrt3+1}\right)^{2n}}{1-\left(\frac{\sqrt3-1}{\sqrt3+1}\right)^{2n}} \end{align} $$ Taking the limit is pretty simple.

Answer 2

HINT:

$\displaystyle\sum_{k=0}^n\binom{2n}{2k}3^k=\sum_{k=0}^n\binom{2n}{2k}(\sqrt3)^{2k}$

and $\displaystyle\sum_{k=0}^{n-1}\binom{2n}{2k+1}3^k=\dfrac{\sum_{k=0}^{n-1}\binom{2n}{2k+1}(\sqrt3)^{2k+1}}{\sqrt3}$

Use $(1+x)^{2n}+(1-x)^{2n}=\cdots$ and $(1+x)^{2n}-(1-x)^{2n}=\cdots$

Answer 3

Hint: You can use that for any $m$, and $i\in\{0,1\}$ one has $$ \sum_{k\equiv i\pmod2}\binom mkX^k =\frac12\sum_k(1^k+(-1)^{k+i})\binom mkX^k =\frac12((1+X)^m+(-1)^i(1-X)^m), $$ and apply this with $m=2n$ and $X=\sqrt3$.

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ ${\ds{\lim_{n\ \to\ \infty}}\ {\ds{\sum_{k\ =\ 0}^{n}\ {2n \choose 2k}3^{k}}\over \ds{\sum_{k\ =\ 0}^{n - 1}\ {2n \choose 2k + 1}3^{k}}}: \ {\large ?}}$.

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The expression $\ds{{2n \choose 2k + s}3^{k}}$, with $\ds{s = 0, 1}$, is 'highly concentrated' around $\ds{\tilde{k}_{s} = {6n - 3s - \root{3} s \over 2\pars{3 + \root{3}}}}$ when $\ds{n \gg 1}$such that $$ \sum_{k\ =\ 0}^{n - s}{2n \choose 2k + s}3^{k} \sim {2n \choose 2\tilde{k}_{s} + s }3^{\tilde{k}_{s}} \int_{0}^{n - s} \exp\pars{-\,{\bracks{k - \tilde{k}_{s}}^{2} \over 2\sigma^{2}}}\,\dd k\,, \qquad\sigma\equiv{\root{3} \over 2\root{3 + 2\root{3}}}\,n^{1/2} $$

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Then, \begin{align} {\ds{\sum_{k\ =\ 0}^{n}\ {2n \choose 2k}3^{k}}\over \ds{\sum_{k\ =\ 0}^{n - 1}\ {2n \choose 2k + 1}3^{k}}} \sim{\ds{2n \choose 2\tilde{k}_{0}}3^{\tilde{k}_{0}} \over \ds{2n \choose 2\tilde{k}_{1} + 1}3^{\tilde{k}_{1}}} =3^{\tilde{k}_{0} - \tilde{k}_{1}}=\root{3} \qquad n \gg 1 \end{align} and \begin{align}\color{#66f}{\large\lim_{n\ \to\ \infty}\ {\ds{\sum_{k\ =\ 0}^{n}\ {2n \choose 2k}3^{k}}\over \ds{\sum_{k\ =\ 0}^{n - 1}\ {2n \choose 2k + 1}3^{k}}}} =\color{#66f}{\large\root{3}} \end{align}