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I have an inifte series and want to show that for any $\alpha >0$ (not necessarily integer) $$\sum_{k=1}^{\infty}\prod_{i=1}^{k}\frac{\alpha-i}{i!} > -1$$ holds. A ratio test yields that the sum is convergent. Additionally, the individual summands converge to $0$ and have their peaks at $k=1$ if $0 <\alpha <1$, at $k=2$ if $1<\alpha <2$ and so forth. However, I cannot figure out any bound.

Thank you in advance.

Phillip
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1 Answers1

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Using binomial theorem for any index :

$$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\dots\infty$$

put $n=\alpha -1$

$$(1+x)^{\alpha-1} = 1+ ({\alpha-1})x + \frac{({\alpha}-1)({\alpha}-2)}{2!}x^2+ \dots \infty$$

Put $x=1$

$$2^{\alpha-1} = 1+ \frac{(\alpha-1)}{1!} + \frac{({\alpha}-1)({\alpha}-2)}{2!} + \dots \infty = 1 +\sum_{k=1}^{\infty}\prod_{i=1}^{k}\frac{\alpha-i}{i!} $$

Since $2^{\alpha -1}$ is always positive for $\alpha \in \mathbb R$ ;

$$ 2^{\alpha -1} =\sum_{k=1}^{\infty}\prod_{i=1}^{k}\frac{\alpha-i}{i!} +1 > 0$$

$$\implies \sum_{k=1}^{\infty}\prod_{i=1}^{k}\frac{\alpha-i}{i!} > -1$$

Jaideep Khare
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  • Nice one. Do you have any quick argument about the convergence of the binomial serie at $x=1$ ? – yultan Mar 23 '17 at 16:44
  • @yultan , I don't know proof of it's convergence.But you can verify it on Wikipedia -> https://en.wikipedia.org/wiki/Binomial_series#Conditions_for_convergence – Jaideep Khare Mar 23 '17 at 16:55