$\lim\limits_{n\rightarrow \infty }\sum\limits_{k=0}^{n}\frac{(-1)^{k}\binom{n}{k}}{2k+1}=0$ Prove that $$\lim_{n\rightarrow \infty }\sum_{k=0}^{n}\frac{(-1)^{k}\binom{n}{k}}{2k+1}=0$$
If $I_{n}=\int_{-1}^{1}(1-x^{2})^{n}dx$, I proved that $(2n+1)I_{n}=2nI_{n-1}$ and I tried to use this result.
 A: $$\begin{aligned}
& \lim_{n\rightarrow \infty } \sum_{k=0}^{n}\frac{(-1)^{k}\binom{n}{k}}{2k+1} \\
=& \lim_{n\rightarrow \infty }\sum_{k=0}^{n}(-1)^{k}\binom{n}{k} \int_0^1 x^{2k} dx \\
=& \lim_{n\rightarrow \infty } \int_0^1\sum_{k=0}^{n}(-1)^{k}\binom{n}{k} x^{2k} dx \\
=& \lim_{n\rightarrow \infty } \int_0^1 (1-x^2)^n dx
\end{aligned}$$
Since when $x \in (0,1)$, $(1-x^2)^n \le 1-x^2$, which is integrable, and $(1-x^2)^n \to 0$ pointwisely as $n\to\infty$, by Lebesgue's Dominated Convergence Theorem,
$$\lim_{n\rightarrow \infty } \int_0^1 (1-x^2)^n dx = \int_0^1 0dx = 0. $$
A: From the Melzak's identity with $f\left(x\right)\equiv1$ and $y=1/2$ we have $$\sum_{k=0}^{n}\dbinom{n}{k}\frac{\left(-1\right)^{k}}{2k+1}=\frac{1}{\dbinom{n+\frac{1}{2}}{n}}\underset{n\rightarrow\infty}{\longrightarrow}0$$ where the limit can be calculated using the Stirling's approximation.
A: One may also avoid the DCT or Stirling's approximation and just notice that
$$ A_n=\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{2k+1}=\int_{0}^{1}(1-x^2)^n\,dx \leq \int_{0}^{1} e^{-nx^2}\,dx \leq \int_{0}^{+\infty} e^{-n x^2}\,dx = \frac{\sqrt{\pi}}{2\sqrt{n}}\to 0.$$
