I want to know is there any property due to which we could write $$\frac{\Gamma(M+\frac{2}{\alpha})}{\Gamma(M)}=\frac{2}{\alpha}\sum_{k=1}^{M} \frac{\Gamma(k-1+\frac{2}{\alpha})}{(k-1)!}$$ Thanks in advance.
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1The tag [tag:proof-verification] should only be used when you're actually verifying a proof is correct - but you haven't included any work or proof here. Also, please use a good title that describes your problem. The title you have communicates absolutely nothing about your question. – Jul 28 '16 at 06:10
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@T.Bongers I have changed the title of the question. – Frank Moses Jul 28 '16 at 06:20
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1Actually I am trying to get the eq(43) of http://arxiv.org/abs/1405.2013 (where the value of $k_1$ is provided in eq(18) of the paper) this equation becomes valid only when the above property is true. Note that I changed the upper limit from $\infty$ to $M$. – Frank Moses Jul 28 '16 at 06:33
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Use $\Gamma(x+1) = x \cdot \Gamma(x)$ and induction. If $M=1$ then $$\frac{\Gamma(1+\frac{2}{\alpha})}{\Gamma(1)}=\Gamma\left(1+\frac{2}{\alpha}\right) =\frac{2}{\alpha}\Gamma\left(\frac{2}{\alpha}\right)=\frac{2}{\alpha}\sum_{k=1}^{1} \frac{\Gamma(k-1+\frac{2}{\alpha})}{(k-1)!}.$$
If $M>1$ then $$\frac{2}{\alpha}\sum_{k=1}^{M} \frac{\Gamma(k-1+\frac{2}{\alpha})}{(k-1)!}= \frac{2}{\alpha}\sum_{k=1}^{M-1} \frac{\Gamma(k-1+\frac{2}{\alpha})}{(k-1)!}+ \frac{2}{\alpha}\frac{\Gamma(M-1+\frac{2}{\alpha})}{(M-1)!}\\= \frac{\Gamma(M-1+\frac{2}{\alpha})}{\Gamma(M-1)}+\frac{2}{\alpha}\frac{\Gamma(M-1+\frac{2}{\alpha})}{(M-1)!}=\left(M-1+\frac{2}{\alpha}\right)\frac{\Gamma(M-1+\frac{2}{\alpha})}{(M-1)!}=\frac{\Gamma(M+\frac{2}{\alpha})}{\Gamma(M)},$$ where we used the induction step and the fact that $\Gamma(n) = (n-1)!$ for any positive integer $n$.
Robert Z
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Your answer is right but is there any other way that we can find that the property is correct. I am a bit skeptic about induction. Nevertheless, it was excellent help and thanks alot. – Frank Moses Jul 28 '16 at 09:18