May not be a research-level problem for an expert, but non-trivial for a non-expert: why do we have
 $$ \sum\_{k=0}^n \frac{(-1)^k}{2k+1} \binom{n}{k} = \frac12 \int\_0^\pi (\sin x)^{2n+1} dx $$
and what is the asymptotic / good lower bound for this as $n$ grows? Thanks!



A relation between a binomial sum and a trigonometric integral
You suspected right. It's easy. Convert $\sin^{2n+1}x=(1-\cos^2x)^n\sin x$ and integrate by substitution and apply binomial expansion
$$\int\_0^{\pi}\sin^{2n+1}x\,dx=2\int\_0^1(1-u^2)^ndu=2\sum\_{k=0}^n(-1)^k\binom{n}k\int\_0^1u^{2k}du=2\sum\_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n}k.$$
There is more to this: the RHS has a closed form
$$\frac12\int\_0^1\sin^{2n+1}x\,dx=\sum\_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n}k=\frac{2^{2n}}{2n+1}\binom{2n}n^{-1}.\tag1$$
To prove (1), one may follow Ira Gessel's suggestion. But, let's employ Zeilberger's algorithm for the [Wilf-Zeilberger method](https://en.wikipedia.org/wiki/Wilf%E2%80%93Zeilberger_pair). Introduce the functions $F(n,k)=(-1)^k\frac{a}{k+a}\binom{n}k\binom{n+a}a$ and $G(n,k)=-F(n,k)\frac{(k+a)k}{(n-k+1)(n+1)}$. Then, check that
$$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k). \tag2$$
Now, sum (2) over all integers (note that $\binom{n}k=0$ if $k<0$ or $k>n$). Then the key is the RHS of (2) vanishes. Further, $f(n+1)-f(n)=0$ where $f(n)=\sum\_{k=0}^nF(n,k)$. So, $f(n)$ is a constant sequence. Checking $f(0)=1$ implies $f(n)=1$, identically. We have recovered the generalized identity that Gessel mentioned above:
$$\sum\_{k=0}^n(-1)^k\frac{a}{k+a}\binom{n}k=\binom{n+a}a^{-1}.$$
**NOTE.** If you're into this kind of integrals, perhaps [this MO discussion](https://mathoverflow.net/questions/253703/integral-of-a-sin-omial-coefficients-binomial) might appeal to you.


Regarding the asymptotic, it can be extracted using [Stirling's approximation](https://en.wikipedia.org/wiki/Stirling's_approximation) for the factorial $n!$; that is, to the first order
$$\frac{2^{2n}}{2n+1}\binom{2n}n^{-1}\,\sim\, \sqrt{\pi}\left(2\sqrt{n}+\frac1{\sqrt{n}}\right)^{-1}.$$


We can get "bonus identities" by considering different linear change of variables for the same integral.
$$\int\_0^1(1-u)^n(1+u)^ndu=\sum\_{k,j=0}^n(-1)^k\binom{n}k\binom{n}j\int\_0^1u^{k+j}du=\sum\_{k,j=0}^n\frac{(-1)^k}{k+j+1}\binom{n}k\binom{n}j.$$
$$\int\_0^1y^n(2-y)^ndy=\sum\_{k=0}^n(-1)^k2^{n-k}\binom{n}k\int\_0^1y^kdy
=\sum\_{k=0}^n\frac{(-1)^k2^{n-k}}{n+k+1}\binom{n}k.$$
$$\int\_{-1}^12^{2n}t^n(1-t)^ndt=2^{2n}\sum\_{k=0}^n(-1)^k\binom{n}k\int\_0^1t^{n+k}dt=2^{2n}\sum\_{k=0}^n\frac{(-1)^k}{n+k+1}\binom{n}k.$$


