Mathematica shows that the infinite series
\begin{align}\sum _{i=0}^{\infty } \rho ^i \prod _{j=1}^i \left(\frac{\alpha }{j}+1\right)= -\frac{(1-\rho )^{-\alpha }}{\rho -1}. \end{align}
How do I prove this?
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$\begingroup$ Starting from the rhs, using the binomial expansion and long division would show it. But, I suppose that you want to start from the lhs, isn't it ? $\endgroup$– Claude LeiboviciCommented Jan 16, 2015 at 11:04
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1 Answer
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Assuming the series converges, we have
\begin{align} & \sum_{i = 0}^\infty \rho^i \prod_{j = 1}^i \left(\frac{\alpha}{j} + 1\right)\\ &= \sum_{i = 0}^\infty \rho^i \prod_{j = 1}^i \frac{\alpha + j}{j}\\ &= \sum_{i = 0}^\infty \rho^i \frac{(\alpha + 1)(\alpha + 2)\cdots (\alpha + i)}{i!}\\ &= \sum_{i = 0}^\infty (-1)^i \rho^i \frac{(-\alpha - 1)(-\alpha - 2)\cdots (-\alpha - i)}{i!}\\ &= \sum_{i = 0}^\infty (-\rho)^i \binom{-\alpha-1}{i}\\ &= (1-\rho)^{-\alpha-1}\\ &= \frac{(1-\rho)^{-\alpha}}{1-\rho}. \end{align}
