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Let's consider more generally the sums (from $0$ to $n$, resp. from $0$ to $n-1$) with a real or complex number $x$ in place of the $n$ inside the sums:

$$\sum_{k=0}^{n}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k}$$ and $$\sum_{k=0}^{n-1}(-1)^k\binom{x+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}\ .$$

It

These sums can be shown evaluated by induction that these sums areas:

$$(-1)^n\frac{n+1}{x(x+1)} \binom{x+n+1}{2n+2}\binom{2n+2}{n+1}$$ and (-1)^n\frac{n+1}{x^2-1} \binom{x+n+1}{2n+3}\binom{2n+4}{n+2} binom{x+n+1}{2n+3}\binom{2n+4}{n+2}\ . We can write these expression respectively as

\begin{equation} \frac{1}{x+1}\left(1+\frac{x}{n+1} \right)\prod_{k=1}^n\left(1-\frac{x^2}{k^2}\right) \end{equation} and \begin{equation} -\frac{2x}{x^2-1}\frac{n+1}{n+2}\prod_{k=1}^{n+1}\left(1-\frac{x^2}{k^2}\right)\ . \end{equation}

The limit of these expression can be evaluated by means of the Euler's infinite product for $\sin(\pi x)$, obtaining respectively the values of the series that you gave:

$$\sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k}=\frac{\sin(\pi x)}{\pi x(x+1)}$$ and $$\sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}=-\frac{2\sin(\pi x)}{\pi(x^2-1)}\ .$$ As observed in the comments, for a positive integer $x=n$, these series coincide respectively with the initially considered finite sums $f$ and $g$, of which they may be considered therefore as a natural extension to complex values of $x$. I will edit and add the details if needed.

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Let's consider more generally the sums (from $0$ to $n$, resp. from $0$ to $n-1$) with a real or complex number $x$ in place of the $n$ inside the sums:

$$\sum_{k=0}^{n}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k}$$ and $$\sum_{k=0}^{n-1}(-1)^k\binom{x+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}\ .$$

It can be shown by induction that these sums are:

$$(-1)^n\frac{n+1}{x(x+1)} \binom{x+n+1}{2n+2}\binom{2n+2}{n+1}$$ and $$(-1)^n\frac{n+1}{x^2-1} \binom{x+n+1}{2n+3}\binom{2n+4}{n+2}$$ The limit of these expression can be evaluated by means of the Euler's infinite product for $\sin(\pi x)$, obtaining respectively the values of the series that you gave:

$$\sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k}=\frac{\sin(\pi x)}{\pi x(x+1)}$$ and $$\sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}=-\frac{2\sin(\pi x)}{\pi(x^2-1)}\ .$$ As observed in the comments, for a positive integer $x=n$, these series coincide respectively with the initially considered finite sums $f$ and $g$, of which they may be considered therefore as a natural extension to complex values of $x$. In case, I will edit later and add the details if needed.

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Warning: the following computations seem to be confirmed by Maple, though they are now a bit fishy to me. I leave them for reference.

Let's consider more generally the sums (from $0$ to $n$, resp. from $0$ to $n-1$) with a real or complex number $x$ in place of the $n$ inside the sums:

$$\sum_{k=0}^{n}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k}$$ and $$\sum_{k=0}^{n-1}(-1)^k\binom{x+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}\ .$$

According to Maple,

It can be shown by induction that these sums toare:

$$(-1)^n\frac{n+1}{x(x+1)} \binom{x+n+1}{2n+2}\binom{2n+2}{n+1}$$ and $$(-1)^n\frac{n+1}{x^2-1} \binom{x+n+1}{2n+3}\binom{2n+4}{n+2}$$ The limit of these expression can be evaluated by means of the Euler's infinite product for $\sin(\pi x)$, obtaining respectively the values of the series that you gave:

$$\sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k} \frac{1}{k+1}\binom{2k}{k}=\frac{\sin(\pi x)}{\pi x(x+1)}$$ and $$\sum_{k=0}^{\infty}(-1)^k\binom{x+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}=-\frac{2\sin(\pi x)}{\pi(x^2-1)}\ .$$ As observed in the comments, for a positive integer $x=n$, these series coincide respectively with the initially considered finite sums $f$ and $g$, of which they may be considered therefore as a natural extension to complex values of $x$. In case, I will edit later and add the details.

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