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Is there any closed form formula (or an equivalent) for this binomial infinite series :

$$F_k(x)= \sum_{n=0}^{\infty} \binom{2n+k}{ n } x^{2n+k} $$ in which |x|<1 and k is a given integer ?

This is a sum over a (odd or even) infinitely long column of Pascal's triangle : enter image description here For example, above : $$F_3(x)= x^3+ 5 x^5 + 21 x^7 + 84 x^9 +... $$

$F_k(x)$ appears in the Fourier Series development of $\frac{1}{1+a\,sin(x)}$ : the k-th harmonic would be of amplitude $F_k(a)$

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  • $\begingroup$ So you are coming from the Chebyshev like relation of $sin^n(x)$ and $sin(nx)$? $\endgroup$
    – Phicar
    Nov 3, 2021 at 17:07

1 Answer 1

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We have $$ S(k,x) = \sum\limits_{n = 0}^\infty {\left( \matrix{ 2n + k \cr n \cr} \right)x^{2n + k} } = x^k \sum\limits_{n = 0}^\infty {\left( \matrix{ 2n + k \cr n \cr} \right)\left( {x^2 } \right)^n } $$

Indicating the series coefficients as $$ t_n = \left( \matrix{ 2n + k \cr n \cr} \right) $$

then we have $$ \eqalign{ & t_0 = 1 \cr & {{t_{n + 1} } \over {t_n }} = {{\left( \matrix{ 2n + 2 + k \cr n + 1 \cr} \right)} \over {\left( \matrix{ 2n + k \cr n \cr} \right)}} = {{n!\left( {n + k} \right)!\left( {2n + 2 + k} \right)!} \over {\left( {n + 1} \right)!\left( {n + 1 + k} \right)!\left( {2n + k} \right)!}} = {{\left( {2n + 2 + k} \right)\left( {2n + 1 + k} \right)} \over {\left( {n + 1} \right)\left( {n + 1 + k} \right)}} = \cr & = 4{{\left( {n + 1 + k/2} \right)\left( {n + 1/2 + k/2} \right)} \over {\left( {n + 1} \right)\left( {n + 1 + k} \right)}} \cr} $$

which means that the series might be expressed through a Hypergeometric function $$ \eqalign{ & S(k,x) = x^k \sum\limits_{0\, \le \;k} {{{\left( {1/2 + k/2} \right)^{\,\overline {\,n\,} } \left( {1 + k/2} \right)^{\,\overline {\,n\,} } } \over {\left( {1 + k} \right)^{\,\overline {\,n\,} } }} {{\left( {4x^2 } \right)^{\,n} } \over {n!}}} = \cr & = x^k {}_2F_{\,1} \left( {\left. {\matrix{ {1/2 + k/2,\;1 + k/2} \cr {1 + k} \cr } \;} \right|\;4x^2 } \right) \cr} $$ which converges for $|x|<1/2$

To explain the convergence range, it is known that $$ S(0,x) = \sum\limits_{n = 0}^\infty {\left( \matrix{ 2n \cr n \cr} \right)x^{2n} } = {1 \over {\sqrt {1 - 4x^2 } }} $$ The duplication formula for Gamma in fact gives $$ \eqalign{ & \left( \matrix{ 2n \cr n \cr} \right) = {{\Gamma \left( {2n + 1} \right)} \over {\Gamma \left( {n + 1} \right)^2 }} = {{4^{\,n} } \over {\sqrt \pi }}{{\Gamma \left( {n + 1/2} \right)} \over {\Gamma \left( {n + 1} \right)}} = \cr & = 4^{\,n} {{\Gamma \left( {n + 1/2} \right)} \over {\Gamma \left( {1/2} \right)\Gamma \left( {n + 1} \right)}} = 4^{\,n} \left( \matrix{ n - 1/2 \cr n \cr} \right) = \cr & = \left( { - 4} \right)^{\,n} \left( \matrix{ - 1/2 \cr n \cr} \right) \cr} $$

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  • $\begingroup$ Wow... impressive. Strange that the convergence is for 1/2, not for 1... counter intuitive $\endgroup$
    – al4085
    Nov 3, 2021 at 18:38
  • $\begingroup$ added explanation of where the $1/2$ comes from $\endgroup$
    – G Cab
    Nov 3, 2021 at 19:04

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