I recently managed to solve a question that I asked a few days ago, but that was unfortunately closed, probably because it wasn't clear enough. Here is the question :
Let $ n $ be a positive integer, prove the following identity : $$ \sum_{k=0}^{n}{\frac{1}{2k+1}\binom{n}{k}}=\left(\sum_{k=0}^{n}{\frac{1}{2^{k}}\binom{2k}{k}}\right)\left(\sum_{k=0}^{n}{\frac{\left(-1\right)^{k}}{2k+1}\binom{n}{k}}\right) $$
What I'm asking you for is to verify the following solution for me.
First of all : $$ \sum_{k=0}^{n}{\frac{1}{2k+1}\binom{n}{k}}=\sum_{k=0}^{n}{\binom{n}{k}\int_{0}^{1}{x^{2k}\,\mathrm{d}x}}=\int_{0}^{1}{\left(1+x^{2}\right)^{n}\,\mathrm{d}x}\overset{\mathrm{denoted}}{=}I_{n}$$
Then :\begin{aligned} I_{n}=\int_{0}^{1}{\left(1+x^{2}\right)^{n}\,\mathrm{d}x}&=\left[x\left(1+x^{2}\right)^{n}\right]_{0}^{1}-2n\int_{0}^{1}{x^{2}\left(1+x^{2}\right)^{n}\,\mathrm{d}x}\\ &=2^{n}-2n\int_{0}^{1}{\left(\left(1+x^{2}\right)^{n}-\left(1+x^{2}\right)^{n-1}\right)\mathrm{d}x} \\ I_{n}&=2^{n}-2n\left(I_{n}-I_{n-1}\right)\\ \iff I_{n}&=\frac{2^{n}}{2n+1}+\frac{2n}{2n+1}I_{n-1}\end{aligned}
In order to solve this recurrence relation, we'll multiply everything by $ \prod\limits_{k=1}^{n}{\frac{2k+1}{2k}} $, so that we could have telescopic cancelling between the consecutive terms. Meaning, $$ \prod_{k=1}^{n}{\frac{2k+1}{2k}}I_{n}=\frac{2^{n}}{2n+1}\prod_{k=1}^{n}{\frac{2k+1}{2k}}+\prod_{k=1}^{n-1}{\frac{2k+1}{2k}}I_{n-1} $$
Since : $$ \frac{2^{n}}{2n+1}\prod_{k=1}^{n}{\frac{2k+1}{2k}}=\frac{2^{n}\prod\limits_{k=0}^{n-1}{\left(2k+1\right)}}{\prod\limits_{k=1}^{n}{\left(2k\right)}}=\frac{\left(2n\right)!}{2^{n}\left(n!\right)^{2}}=\frac{1}{2^{n}}\binom{2n}{n} $$
We get that for any $ k\geq 1 $ : \begin{aligned} \prod_{j=1}^{k}{\frac{2j+1}{2j}}I_{k}-\prod_{j=1}^{k-1}{\frac{2j+1}{2j}}I_{k-1}&=\frac{1}{2^{k}}\binom{2k}{k}\\ \Longrightarrow\sum_{k=1}^{n}{\left(\prod_{j=1}^{k}{\frac{2j+1}{2j}}I_{k}-\prod_{j=1}^{k-1}{\frac{2j+1}{2j}}I_{k-1}\right)}&=\sum_{k=1}^{n}{\frac{1}{2^{k}}\binom{2k}{k}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \prod_{k=1}^{n}{\frac{2k+1}{2k}}I_{n}-1&=\sum_{k=1}^{n}{\frac{1}{2^{k}}\binom{2k}{k}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I_{n}&=\left(\prod_{k=1}^{n}{\frac{2k}{2k+1}}\right)\left(\sum_{k=0}^{n}{\frac{1}{2^{k}}\binom{2k}{k}}\right)\end{aligned}
Secondly : $$ \sum_{k=0}^{n}{\frac{\left(-1\right)^{k}}{2k+1}\binom{n}{k}}=\sum_{k=0}^{n}{\left(-1\right)^{k}\binom{n}{k}\int_{0}^{1}{x^{2k}\,\mathrm{d}x}}=\int_{0}^{1}{\left(1-x^{2}\right)^{n}\,\mathrm{d}x}\overset{\mathrm{denoted}}{=}J_{n} $$
Then :\begin{aligned} J_{n}=\int_{0}^{1}{\left(1-x^{2}\right)^{n}\,\mathrm{d}x}&=\left[x\left(1-x^{2}\right)^{n}\right]_{0}^{1}+2n\int_{0}^{1}{x^{2}\left(1-x^{2}\right)^{n}\,\mathrm{d}x}\\&=2n\int_{0}^{1}{\left(\left(1-x^{2}\right)^{n-1}-\left(1-x^{2}\right)^{n}\right)\mathrm{d}x} \\ J_{n}&=2n\left(J_{n-1}-J_{n}\right)\\ \iff \ \ \ \ \ \ \ \ \ J_{n}&=\frac{2n}{2n+1}J_{n-1}\\ \Longrightarrow \prod_{k=1}^{n}{\frac{J_{k}}{J_{k-1}}}&=\prod_{k=1}^{n}{\frac{2k}{2k+1}}\\ \iff \ \ \ \ \ \ \ \ \ J_{n}&=\prod_{k=1}^{n}{\frac{2k}{2k+1}}\end{aligned}
Thus, $$ \sum_{k=0}^{n}{\frac{1}{2k+1}\binom{n}{k}}=\left(\sum_{k=0}^{n}{\frac{\left(-1\right)^{k}}{2k+1}\binom{n}{k}}\right)\left(\sum_{k=0}^{n}{\frac{1}{2^{k}}\binom{2k}{k}}\right) $$
What do you guys think ?