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I'm inerested in the asymptotic behaviour of

$$ f_n(2,2) := \sum_{k=0}^n \begin{pmatrix} 2n \\ k \end{pmatrix} 2^k $$

for $n\to\infty$. More precisely, I want to find the base $b$ such that for any $\epsilon > 0$ the following holds:

$$ \frac{f_n(2,2)}{(b+\epsilon)^n} \to 0 \quad \text{and} \quad \frac{f_n(2,2)}{(b-\epsilon)^n} \to \infty.$$

Approach: There is this generating function for the binomial coefficients $$ f_n(1,x) := (1+x)^n = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^k. $$

Is there maybe something similar for this generalized function $f_n(c,x)$ with $$f_n(c,x) := \sum_{k=0}^n \begin{pmatrix} cn \\ k \end{pmatrix} x^k ?$$ Or does, just, anybody have an idea how to get the base $b$ for $f_n(2,2)$?

Thank you

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Since the terms of the sum $\sum_{k=0}^{n}\binom{2n}{k}2^k $ are rapidly increasing, it makes sense to perform a reindexing $k\mapsto n-k$ and to write such sum as $$ 2^n\sum_{k=0}^{n}\binom{2n}{n-k}\frac{1}{2^k}=2^n\binom{2n}{n}\sum_{k=0}^{n}\frac{n!^2}{2^k(n-k)!(n+k)!}=2^n\binom{2n}{n}\cdot\phantom{}_2 F_1(1,-n;n+1,-\tfrac{1}{2}) $$ where the last hypergeometric function is convergent to $2$ as $n\to +\infty$ by the dominated convergence theorem. In particular $$ \sum_{k=0}^{n}\binom{2n}{k}2^k \sim \frac{2^{3n+1}}{\sqrt{\pi n}} $$ as $n\to +\infty$, hence $b=\color{red}{8}$. A more accurate approximation (enclosing the second term of the asymptotic expansion) is $\frac{2^{3n+1}}{\sqrt{\pi n}\left(1+\frac{3}{n}\right)}$.

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  • $\begingroup$ Thank you very much. This was very helpful. $\endgroup$ – diddy Jun 21 '18 at 13:54
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Probably not very useful.

As Jack D'Aurizio showed in his answer, you are entering the world of hypergeometric functions and $$f_n(c,x) = \sum_{k=0}^n \begin{pmatrix} c\,n \\ k \end{pmatrix} x^k =(x+1)^{c \,n}-x^{n+1} \binom{c\, n}{n+1} \, _2F_1(1,-c\,n+n+1;n+2;-x)$$

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