query: How to derive the close form of a power series with a ${2n\choose n}$ binomial coefficient?
positive_passages.docid: 27588
positive_passages.text: Generating functions and central binomial coefficient
negative_passages.docid: 43188
negative_passages.text: The $n^{th}$ root of the geometric mean of binomial coefficients.
negative_passages.docid: 140462
negative_passages.text: Find a closed form for the power series $f_p(x)=\sum\limits_{n=0}^\infty\frac{n^p}{n!}x^n$
negative_passages.docid: 83346
negative_passages.text: Finding a closed form $x_n$ for the recurrence relation $x_{n+2} = px_{n+1} + qx_n + r$
negative_passages.docid: 151384
negative_passages.text: Summation of series involving binomial coefficients and polynomial of degree at most n-1
negative_passages.docid: 69084
negative_passages.text: What is the closed form for $\sum_{n=1}^\infty \frac1n - \frac1{n+1/p}$?
negative_passages.docid: 51509
negative_passages.text: How to show that $1 \over \sqrt{1 - 4x} $ generates $\sum_{n=0}^\infty \binom{2n}{n}x^n $
negative_passages.docid: 131468
negative_passages.text: Summation of Central Binomial Coefficients divided by even powers of $2$
negative_passages.docid: 108401
negative_passages.text: Sum of binomial coefficients $\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2n - 2k}{n - 1} = 0$
negative_passages.docid: 136652
negative_passages.text: Closed form expression for the product $\prod\limits_{k=1}^{n}\left(1 - \frac{1}{ak}\right)$
negative_passages.docid: 139307
negative_passages.text: Closed form expression for infinite series
negative_passages.docid: 42606
negative_passages.text: Orthogonality for Binomial Coefficients
negative_passages.docid: 47052
negative_passages.text: How can I simplify this expression involving binomial coefficients?
negative_passages.docid: 59253
negative_passages.text: Summing the power series $\sum\limits_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}\prod\limits_{k=1}^n\frac{2k-1}{2k} $
negative_passages.docid: 32791
negative_passages.text: Proving this binomial identity $\sum_{k=0}^n {n+k \choose k} \frac{1}{2^{k}}= 2^{n}$
negative_passages.docid: 71797
negative_passages.text: Combinatorial argument for $\sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i} = \binom{n}{i}2^{n-i}$
negative_passages.docid: 35381
negative_passages.text: Sum with binomial coefficients: $\sum_{k=1}^m  \frac{1}{k}{m \choose k} $