In this question, you need not consider issues of convergence.
The sequence \(T_n\), for \(n = 0, 1, 2, \ldots\), is defined by \(T_0 = 1\) and, for \(n \geqslant 1\), by
\[ T_n = \frac{2n-1}{2n}\,T_{n-1}. \]
Prove by induction that
\[ T_n = \frac{1}{2^{2n}}\binom{2n}{n}, \]
for \(n = 0, 1, 2, \ldots\).
[Note that \(\dbinom{0}{0} = 1\).]
Show that in the binomial series for \((1-x)^{-\frac{1}{2}}\),
\[ (1-x)^{-\frac{1}{2}} = \sum_{r=0}^{\infty} a_r x^r, \]
successive coefficients are related by
\[ a_r = \frac{2r-1}{2r}\,a_{r-1} \]
for \(r = 1, 2, \ldots\)\,.
Hence prove that \(a_r = T_r\) for all \(r = 0, 1, 2, \ldots\)\,.
Let \(b_r\) be the coefficient of \(x^r\) in the binomial series for \((1-x)^{-\frac{3}{2}}\), so that
\[ (1-x)^{-\frac{3}{2}} = \sum_{r=0}^{\infty} b_r x^r. \]
By considering \(\dfrac{b_r}{a_r}\), find an expression involving a binomial coefficient for \(b_r\), for \(r = 0, 1, 2, \ldots\)\,.
By considering the product of the binomial series for \((1-x)^{-\frac{1}{2}}\) and \((1-x)^{-1}\), prove that
\[ \frac{(2n+1)}{2^{2n}}\binom{2n}{n} = \sum_{r=0}^{n} \frac{1}{2^{2r}}\binom{2r}{r}, \]
for \(n = 1, 2, \ldots\)\,.
Solution:
Claim: \(\displaystyle T_n = \frac{1}{2^{2n}}\binom{2n}{n}\)
Proof: (By Induction)
Base case: \(n=0\). Note that \(T_0 = 1\) and \(\frac{1}{2^0}\binom{0}{0} = 1\) so the base case is true.
Assume true for some \(n=k\), ie \(T_k = \frac{1}{2^{2k}} \binom{2k}{k}\) so
\begin{align*}
&& T_{k+1} &= \frac{2(k+1)-1}{2(k+1)} \frac{1}{2^{2k}} \binom{2k}{k} \\
&&&= \frac{2k+1}{k+1} \frac{1}{2^{2k+1}} \frac{(2k)!}{k!k!} \\
&&&= \frac{2(k+1)(2k+1)}{(k+1)(k+1)} \frac{1}{2^{2(k+1)}} \frac{(2k)!}{k!k!} \\
&&&= \frac{1}{2^{2(k+1)}} \frac{(2k+2)!}{(k+1)!(k+1)!} \\
&&&= \frac{1}{2^{2(k+1)}} \binom{2(k+1)}{k+1}
\end{align*}
and therefore it's true for all \(n\).
Notice that \((1-x)^{-\frac12} = 1 + (-\tfrac12)(-x) + \frac{(-\frac12)(-\frac32)}{2!}(-x)^2+\cdots\) in particular \(a_r = \frac{(-\frac12 - r)}{r}(-1)a_{r-1} = \frac{2r-1}{2r}a_{r-1}\). Since \(a_0 = 1\) we have \(a_r = T_r\) for all \(r\).
Notice that \begin{align*}
&& (1-x)^{-\frac32} &= (1-x)^{-\frac12}(1-x)^{-1} \\
&&&= (1 + x+ x^2 + \cdots) \sum_{r=0}^{\infty} a_r x^r \\
&&&= \sum_{i=0}^{\infty} \sum_{k=0}^n a_r x^i
\end{align*}
So we must have \(b_r = \sum_{i=0}^ra_i\) which is the required result