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Given the following sum:

$$0.5\cdot\sum\limits_{k=0}^\infty \frac{1}{k+1}\binom{2k}{k}\cdot(0.25)^{k}$$ I know that the sum is supposed to converge to $1$. How would I go about evaluating it to get this result? I thought about binomial theorem, but the fraction $\frac{1}{k+1}$ makes it rather problematic.

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2 Answers 2

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Let $C_n=\frac1{n+1}\binom{2n}n$ be the $n$-th Catalan number. As derived here, the generating function for the Catalan numbers is

$$c(x)=\sum_{n\ge 0}C_nx^n=\frac{1-\sqrt{1-4x}}{2x}\;.$$

Thus, your expression is simply $\frac12c\left(\frac14\right)$.

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Try $\frac{x^{k+1}}{k+1} = \int_{0}^{x} y^k dy$ and interchange of integral and sum.

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