The background context is this old MSE question here. Essentially I was trying to write up a proper response for the question, but I'm stuck on one part myself. One goal for the question is to prove the following equality:
$$\sum_{k=0}^{n+1} \binom n k \frac{(-1)^k}{(n+k)(n+k+1)} = \int_0^1x^{n-1}(1-x)^{n+1}dx$$
(At least, based on the assumption the OP's notation is that $C_r := \binom n r$ where $n$ is understood to be fixed. Seems to work right when calculating both via Wolfram.) Per a suggestion in the comments, it seems a good route route to go would be expanding $(1-x)^{n+1}$ via the binomial theorem and integrating termwise, and hopefully the expression for the sum "pops out," as it were.
Doing so, we find that
$$\begin{align} \int_0^1x^{n-1}(1-x)^{n+1}dx &= \int_0^1 x^{n-1} \sum_{k=0}^{n+1} \binom {n+1} k (-1)^k x^{k}dx \\ &= \int_0^1 \sum_{k=0}^{n+1} \binom {n+1} k (-1)^k x^{n+k-1}dx \\ &= \sum_{k=0}^{n+1} \binom {n+1} k (-1)^k \int_0^1 x^{n+k-1}dx \\ &= \sum_{k=0}^{n+1} \left. \binom {n+1} k \frac{ (-1)^k}{n+k} x^{n+k} \right|_{x=0}^1\\ &= \sum_{k=0}^{n+1} \binom {n+1} k \frac{ (-1)^k}{n+k} \end{align}$$
So far so good, but with one issue. What remains, seemingly, is to show
$$\binom n k \frac{1}{n+k+1} = \binom{n+1}k \tag 1$$
However, these two quantities are unequal. For instance, $n=10,k=5$ give us $63/4 = 462$, pure nonsense. So this suggests something in my work along the way is wrong ... but what? I've checked and double-checked a lot, and I'm not sure where I went wrong in this derivation, or how to move forward. Perhaps I'm overlooking something obvious, but does anyone have any ideas?
One hypothesis was that the original inequality I seek to prove is false, but it seems not. For instance, take $n=15$. Then the sum and integral evaluate to about $2.2182 \times 10^{-10}$ per WolframAlpha, and I tried a few other $n$ besides that. So the issue almost certainly lies in my derivation. But my sum also evaluates to this for the prescribed value of $n=15$... which suggests another possible problem.
Another possibility is that I'm oversimplifying the matter by assuming
$$\sum_{k=0}^{n+1} \binom n k \frac{(-1)^k}{(n+k)(n+k+1)} = \sum_{k=0}^{n+1} \binom {n+1} k \frac{ (-1)^k}{n+k}$$
would imply corresponding terms are equal (which was my motivation for the earlier equality $(1)$ which is a clear dead-end). For clarity, that means I assumed $\sum a_n = \sum b_n \implies a_n = b_n$. While that'd obviously make everyone's life easier, it's definitely not necessarily true. Which, if it's not in this case, I'm not sure where to go, and would like any potential nudges forward on the matter.
$$\begin{align*} \sum_{k=0}^{n+1}\binom{n}k\frac{(-1)^k}{(n+k)(n+k+1)}&=\sum_{k=0}^{n+1}\binom{n}k\left(\frac{(-1)^k}{n+k}-\frac{(-1)^k}{n+k+1}\right)\\ &=\sum_{k=0}^{n+1}\binom{n}k\frac{(-1)^k}{n+k}-\sum_{k=0}^{n+1}\binom{n}k\frac{(-1)^k}{n+k+1}\\ &=\sum_{k=0}^{n+1}\binom{n}k\frac{(-1)^k}{n+k}+\sum_{k=0}^{n+1}\binom{n}k\frac{(-1)^{k+1}}{n+k+1}\\ &=\sum_{k=0}^{n+1}\binom{n}k\frac{(-1)^k}{n+k}+\sum_{k=1}^{n+1}\binom{n}{k-1}\frac{(-1)^k}{n+k}\\ &=\frac1n+\sum_{k=1}^{n+1}\left(\binom{n}k+\binom{n}{k-1}\right)\frac{(-1)^k}{n+k}\\ &=\frac1n+\sum_{k=1}^{n+1}\binom{n+1}k\frac{(-1)^k}{n+k}\\ &=\sum_{k=0}^{n+1}\binom{n+1}k\frac{(-1)^k}{n+k} \end{align*}$$