A series sum involving Catalan numbers I was trying to compute $$\sum_{k=0}^{n} \left(-\frac{1}{2}\right)^k \, \binom{2k}{k} \, \frac{k}{k+1} = \sum_{k=0}^n \left(-\frac12\right)^k k C_k$$
(where $C_k$ is the $k^{\rm th}$ Catalan number) but could not come up with a good idea.
I found out it is equal to  $$\sum_{k=0}^{n} \left(-\frac{1}{2}\right)^k \, \binom{2k}{k+1}.$$ Doing some computation I see it is also equal to $$\sum_{k=0}^{n} \left(\frac{1}{2}\right)^k \, \binom{-k-1}{k}.$$ If someone can help, I would be very greatful. Thanks... My attempts might be wrong.
 A: The ordinary generating function for your sums is $$ g(x) = \frac{x +1-\sqrt{1+2 x }}{x \left(x -1\right) \sqrt{1+2x}}$$
That is, your sum is the coefficient of $x^n$ in the Maclaurin series of $g(x)$.
A: Using Gosper's algorithm, you can prove that your summation does not have any hypergeometric closed form. Roughly, this means that the summation cannot be expressed as a rational function whose numerator and denominator are a product of polynomials in $n$, factorials (more generally, $\Gamma(an+b)$, for complex constants $a$ and $b$),  and constants to the $n^{th}$ power. This includes anything involving binomial coefficients.
Technically, this does not rule out closed forms involving expressions like $n^n$, or Stirling numbers, but I suspect there really is no "nice" closed form.
This Maxima code outputs NO_HYP_SOL, which confirms my claim. Try it online!
load(zeilberger)$

out: Gosper((-1/2)^k * binomial(2*k,k) * k / (k+1), k)$
fortran(out);

A: One can view the Oeis sequences to find A014138 which is a finite sum of Catalan numbers given as $$ a_{n} = \sum_{k=0}^{n} C_{k}. $$ So there is a base case of the series. Now, consider the series
$$ a_{n}(x) = \sum_{k=0}^{n} C_{k} \, x^k $$
which leads to
$$ x \, \frac{d \, a_{n}(x)}{d x} = b_{n}(x) = \sum_{k=0}^{n} k \, C_{k} \, x^k $$
which have the generating functions
$$ \sum_{n=0}^{\infty} a_{n}(x) \, t^n = \frac{1 - \sqrt{1 - 4 x t}}{2 \, x \, t \, (1-t)} $$
and
$$ \sum_{n=0}^{\infty} x \, \frac{d \, a_{n}(x)}{dx} \, t^n = \frac{1 - 2 x t - \sqrt{1 - 4 x t}}{2 \, x^2 \, t \, (1-t) \, \sqrt{1 - 4 x t}}, $$
respectively. Setting $x = - 1/2$ gives the desired generating function, which has been given in another answer.
Another way to consider the series is the use the integral representation
$$ C_{n} = \frac{2^{2 n +1}}{\pi} \, \int_{-1}^{1} t^{2 n } \, \sqrt{1- t^2} \, dt $$
which leads to
\begin{align}
b_{n}\left(- \frac{1}{2}\right) &= \frac{2}{\pi} \, \int_{-1}^{1} \sqrt{1 - t^2} \, \left( \sum_{k=0}^{n} k \, (- 2 t^2)^k \right) \, dt \\
&= - \frac{4}{\pi} \, \left( I_{1} - (n+1) \, (-2)^n \, I_{2} + n \, (-2)^{n+1} \, I_{3} \right),
\end{align}
where
\begin{align}
I_{1} &= \int_{-1}^{1} \frac{t^2 \, \sqrt{1 - t^2}}{(1 + 2 t^2)^2} \, dt = \frac{2 \sqrt{3} - 3}{12} \\
I_{2} &= \int_{-1}^{1} \frac{t^{2n +2} \, \sqrt{1 - t^2}}{(1 + 2 t^2)^2} \, dt = - \frac{\sqrt{\pi} \, \Gamma\left(n + \frac{3}{2}\right)}{6 \, (n+2)!} \, \left((3 n +2) F_{1} - n - 2 \right) \\
I_{3} &= \int_{-1}^{1} \frac{t^{2 n + 4} \, \sqrt{1 - t^2}}{(1 + 2 t^2)^2} \, dt = - \frac{\sqrt{\pi} \, \Gamma\left(n + \frac{5}{2}\right)}{6 \, (n+3)!} \, \left((3 n +5) F_{2} - n - 3 \right) \\
F_{1} &= {}_{2}F_{1}\left(1, \, n + \frac{3}{2}; \, n+3; \, -2 \right) \\
F_{2} &= {}_{2}F_{1}\left(1, \, n + \frac{5}{2}; \, n+4; \, -2 \right).
\end{align}
This leads to
\begin{align}
b_{n}\left(- \frac{1}{2}\right) &= \frac{3 - 2 \, \sqrt{3}}{3 \, \pi} + \frac{(-1)^n}{3 \cdot 2^{n+2}} \, \left(2 (n+1) \, \binom{2n +2}{n+1} + n \, \binom{2n+4}{n+2} \right) \\
& \hspace{10mm} - \frac{(-2)^{n+2}}{\pi} \, \left((n+1) \, F_{1} - 2 \, n \, F_{2} \right).
\end{align}
