Another was is to use the generating function of the central binomial coefficients, that is $$\frac{1}{\sqrt{1-x}}=\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{4^{n}}x^{n}.$$ We will make use of the fact that $x_{n}=\binom{2n}{n}\frac{1}{4^{n}}$ is a monotically decreasing sequence to make sure there are no oscillations. It is monotonic since $$x_{n+1}=\binom{2n}{n}\frac{1}{4^{n}}\frac{(2n+2)(2n+1)}{4\left(n+1\right)^{2}}=x_{n}\left(1-\frac{1}{2n+2}\right).$$ Taking the indefinite integral of both sides of the generating function, we have that for $|x|<1,$ $$2-2\sqrt{1-x}=\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{4^{n}}\frac{x^{n+1}}{n+1}.$$ Now, the left hand side converges as $x\rightarrow1$ from the left, and so, since the series has strictly positive coefficients, the right hand side must converge as well. This implies that $$\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{4^{n}}\frac{1}{n+1}$$ is a convergent series, and since the coefficients $\binom{2n}{n}\frac{1}{4^{n}}$ are monotonically decreasing, the divergence of the harmonic series implies that we must have $$\binom{2n}{n}\frac{1}{4^{n}}\rightarrow0.$$
Added Details: Based on the comments, I thought I would give a precise proof of why the series must converge, without appealing to Littlewood's Tauberian theorem. The proof only requires that the coefficients of the power series, $a_n$, are nonnegative. Let $a_n\geq0$, $s_m=\sum_{n=0}^m a_n$ be the partial sums, and suppose that $\sum_{n=0}^{\infty}a_{n}x^{n}\leq C$ for all $x\in(0,1),$ for some positive constant $C$. Then for $x\in(0,1),$ $$s_{m}=\sum_{n=0}^{m}a_{n}\left(1-x^{n}\right)+\sum_{n=0}^{m}a_{n}x^{n}$$ $$\leq(1-x)\sum_{n=0}^{m}a_{n}\left(1+x+\cdots+x^{n-1}\right)+C$$
$$\leq(1-x)\sum_{n=0}^{m}na_{n}+C.$$ Taking $x$ sufficiently close to $1$, we can make $(1-x)\sum_{n=0}^{m}na_{n}\leq 1$, and so it follows that $s_{m}\leq C+1.$ This means that $s_m$ is monotonically increasing and bounded, its limit, $\sum_{n=0}^\infty a_n$ must converge. In particular, our series $$\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{4^{n}(n+1)},$$ is convergent.