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For all natural numbers $n$, prove that

$$ \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {2r+1} \right) = 4^n \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1} \right)$$

I have tried to do this by induction, but I'm not getting anywhere with that:

$$ \sum_{r=0}^{k+1} \left( \frac{ (-1)^r { k+1 \choose r} } {2r+1}\right) \\ = 1 + \sum_{r=1}^{k} \left( \frac{ (-1)^r ( { k \choose r} + { k \choose r-1}) } {2r+1}\right) + \frac{(-1)^r}{2k+3} \\ = 4^k \sum_{r=0}^k \left( \frac{ (-1)^r {k \choose r} } {k+r+1} \right) + \sum_{r=1}^{k} \left( \frac{ (-1)^r { k \choose r-1} } {2r+1}\right) + \frac{(-1)^r}{2k+3} $$ ... Which is nowhere close to the desired $4^{k+1} \sum_{r=0}^{k+1} \left( \frac{ (-1)^r {k+1 \choose r} } {k+1+r+1} \right)$

Here are a few verified values, I have tested them up to $n=30$ and they all work. When the numbers get too large, my program gets a little inconsistent with the calculations. But WolframAlpha still manages, when $n=10000$, LHS = RHS $\approx 0.0088619369$.

$$ \begin{array}{|c|c|c|} \hline \text{n}& \text{LHS} & \text{RHS} \\ \hline 0& 1& 1\\ \hline 1& \frac{2}{3}& \frac{2}{3}\\ \hline 2& \frac{8}{15}& \frac{8}{15}\\ \hline 3& \frac{16}{35}& \frac{16}{35}\\ \hline 4& \frac{128}{315}& \frac{128}{315}\\ \hline 5& \frac{256}{693}& \frac{256}{693}\\ \hline 6& \frac{1024}{3003}& \frac{1024}{3003}\\ \hline 7& \frac{2048}{6435}& \frac{2048}{6435}\\ \hline 8& \frac{32768}{109395}& \frac{32768}{109395}\\ \hline 9& \frac{65536}{230945}& \frac{65536}{230945}\\ \hline 10& \frac{262144}{969969}& \frac{262144}{969969}\\ \hline \end{array}$$

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    $\begingroup$ For large $n$ I'm sure any discrepancies are due to numerical phenomena such as rounding error, underflow etc. $\endgroup$ – Lord Shark the Unknown Nov 10 '18 at 18:09
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The left hand side equals $$\sum_{r=0}^n(-1)^r\binom{n}r\int_0^1 x^{2n}\,dx=\int_0^1(1-x^2)^n\,dx.\tag{1}$$ The left hand side equals $$4^n\sum_{r=0}^n(-1)^r\binom{n}r\int_0^1 x^{n+r}\,dx=4^n\int_0^1x^n(1-x)^n\,dx. \tag{2}$$ The integral (1) equals $$\frac12\int_{-1}^1(1-x^2)^n\,dx$$ and if we substitute $y=(1+x)/2$ we get $$\int_0^1(4y(1-y))^n\,dy$$ which is (2).

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We may recall the partial fraction decomposition of the reciprocal Pochhammer symbol:

$$ \frac{1}{x(x+1)(x+2)\cdot\ldots\cdot(x+n)}=\frac{1}{(x)_{n+1}}=\frac{1}{n!}\sum_{k=0}^{n}\frac{(-1)^{k}\binom{n}{k}}{x+k} $$ and simply evaluate both sides at $x=1$ and $x=\frac{1}{2}$.

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