330 followers problem-1

Let $\displaystyle S=\sum_{k=0}^{\infty} \binom{2k+1}{k}x^k = \sum_{k=0}^{\infty} \binom{2k}{k}x^k (\frac{2k+1}{k+1}) = \sum_{k=0}^{\infty}(1-\frac{1}{k+1})\binom{2k}{k}x^k$

So we may write $\displaystyle S = \underbrace{\color{#D61F06}{\sum_{k\ge 0}\binom{2k}{k}x^k}}_{\text{Central Binomial Generating Function}} - \underbrace{\color{#3D99F6}{\sum_{k\ge 0} \frac{1}{k+1}\binom{2k}{k}x^k}}_{\text{Catalan Numbers}} = \color{#333333}{\frac{1}{\sqrt{1-4x}} - \frac{1-\sqrt{1-4x}}{2x}}$

Putting $\displaystyle x=\frac{7}{64}$ we get, $\displaystyle S=\boxed{\frac{32}{21}}$

$\text{Evaluation of : }\color{#3D99F6}{\displaystyle \sum_{n\ge 0}\frac{1}{n+1}\binom{2n}{n}x^n}$

The Catalan Numbers satisfy the recurrence relation $\displaystyle C_n = \sum_{k=0}^{n-1}C_k C_{n-k-1}$

Let $\displaystyle A(x)=\sum_{n\ge 0}C_n x^n = \sum_{n\ge 0}\sum_{k=0}^{n-1}C_k C_{n-k-1} x^n = 1+ \sum_{n\ge 1}\sum_{k=0}^{n-1}C_k C_{n-k-1} x^n=1+ x\sum_{n\ge 0}\sum_{k=0}^{n}C_k C_{n-k-1} x^n$

Now by Cauchy product we know $\displaystyle (A(x))^2 = (\sum_{n\ge 0}C_n x^n)(\sum_{n\ge 0}C_n x^n) = \sum_{n\ge 0}\sum_{k=0}^{n}C_k C_{n-k}x^n$

Therefore, $\displaystyle A(x)=1+x(A(x))^2$ and solving the quadratic we have $\displaystyle A(x) =\sum_{n\ge 0}\frac{1}{n+1}\binom{2n}{n}x^n =\frac{1-\sqrt{1-4x}}{2x}$

The Red sum can be obtained by differentiating this sum.