Let $0 < |x|<1$ and let $$a_n = \sum_{i = n+1}^\infty \frac{i-n}{i+1}\binom{2i}{i}x^i.$$ And I am trying to show that $$ \lim_{n \to \infty} a_n = 0. $$
I am trying to use the following equality: \begin{align} f_n(x) &= \sum_{i = 0}^\infty \frac{(i-n)}{i+1}\binom{2i}{i}x^i \\ &= \frac{1}{\sqrt{1-4x}} - (n+1)\frac{1-\sqrt{1-4x}}{2x} \end{align} which can be obtained by using LINK. But not sure how to involve this...
Any comments or suggestions will be very appreciated. Thanks in advance.
For $|x|< 1/4$ the sequence converges to $0$.
There is a simple upper bound given by: $$|a_n| \le \sum_{i=n+1}^{\infty} \frac{i-n}{i+1} \binom{2i}{i} |x|^i \le \sum_{i=n+1}^{\infty} \binom{2i}{i} |x|^i$$ since $\frac{i-n}{i+1} <1$.
Now, it is natural to consider the series $$\sum_{i=1}^{\infty} \binom{2i}{i} |x|^i$$ which is convergent by the ratio test (the radius of convergence is $1/4$). Thus, the remainders get infinitesimal $$\sum_{i=n+1}^{\infty} \binom{2i}{i} |x|^i \to 0$$ Giving you $|a_n| \to 0$ as $n \to \infty$.