$$\sum_{k=1}^n\left(\frac{C_{2k-1}}{2k}\right)=\frac{2^n-1}{n+1}$$
I tried decoding that sigma on the LHS to identify the pattern. However, I wasn't able to proceed any further. I tried integrating the expansion of $(1-x)^n$ and its expansion, and then multiplying it with $(x+1)^n$ and its expansion; mainly to check if I am able to simplify anything. But no luck. I need help with this problem. Thanks in advance, for any and all help!
Ok, I'm gonna assume
$$ C_{2k-1} = \binom{n}{2k-1}$$
Consider,
$$ f(x) = (1+x)^n = \sum_{i=0}^n \binom{n}{k} x^i$$
Notice, that odd coefficents are of interest here. We can extract out the odd part of function as:
$$ \frac{ f(x) - f(-x) }{2} = \sum_{k=1}^n x^{2k-1} \binom{n}{2k-1}$$
Now, to get the $ \frac{1}{k}$ part, integrate both sides:
$$ \int_{0}^1 \frac{ (1+x)^n - (1-x)^{n} }{2} dx = \sum_{k=1}^n \frac{1}{2k} \binom{n}{2k-1}$$
I'll leave it upto you to evaluate the LHS :^)
$$\begin{align}\sum_{k=1}^n\frac{\binom n{2k-1}}{2k}&=\sum_{k=1}^{\lfloor(n+1)/2 \rfloor}\frac{\binom n{2k-1}}{2k}\\&= \sum_{k=1}^{\lfloor(n+1)/2 \rfloor}\frac{\binom{n+1}{2k}}{n+1}\\ &=\frac1{n+1}\left(\sum_{k=0}^{\lfloor(n+1)/2 \rfloor}\binom{n+1}{2k}-1\right) \\ &=\frac1{n+1}\left(2^n-1\right), \end{align}$$ since $\sum_{k=0}^{\lfloor m/2 \rfloor}\binom m{2k}=2^{m-1}$.
