Help with complex analysis Watch

I Closed My Eyes
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#1
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I need help proving that for  z \in \mathbb{C} , given

 G_{2,m} (z) := \sum_{k=0}^{\infty} \frac{z^{k}}{\Gamma (k+2) \Gamma (mk+1)} ,

show that
 zG_{2,0} (z) = e^{z} - 1 .


What I've done so far is
 zG_{2,0} (z) = \sum_{k=0}^{\infty} \frac{z^{k+1}}{\Gamma (k+2)} =\sum_{k=0}^{\infty} \frac{z^{k}}{k!} \frac{z}{k+1}.

But I don't know where to take it from there. I would appreciate for some guidance on what to do after this step.

Thanks
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DFranklin
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(Original post by I Closed My Eyes)
I need help proving that for  z \in \mathbb{C} , given

 G_{2,m} (z) := \sum_{k=0}^{\infty} \frac{z^{k}}{\Gamma (k+2) \Gamma (mk+1)} ,

show that
 zG_{2,0} (z) = e^{z} - 1 .


What I've done so far is
 zG_{2,0} (z) = \sum_{k=0}^{\infty} \frac{z^{k+1}}{\Gamma (k+2)} =\sum_{k=0}^{\infty} \frac{z^{k}}{k!} \frac{z}{k+1}.

But I don't know where to take it from there. I would appreciate for some guidance on what to do after this step.

Thanks
Surely this is just the series for e^z with the first term omitted (so = exp(z) - 1), so you're done?

Have to say, seems pretty advanced for year 9...
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RDKGames
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(Original post by I Closed My Eyes)
I need help proving that for  z \in \mathbb{C} , given

 G_{2,m} (z) := \sum_{k=0}^{\infty} \frac{z^{k}}{\Gamma (k+2) \Gamma (mk+1)} ,

show that
 zG_{2,0} (z) = e^{z} - 1 .


What I've done so far is
 zG_{2,0} (z) = \sum_{k=0}^{\infty} \frac{z^{k+1}}{\Gamma (k+2)} =\sum_{k=0}^{\infty} \frac{z^{k}}{k!} \frac{z}{k+1}.

But I don't know where to take it from there. I would appreciate for some guidance on what to do after this step.

Thanks
To add to the above, note that \dfrac{z^k}{k!} \cdot \dfrac{z}{k+1} = \dfrac{z^{k+1}}{(k+1)!} from which it should be obvious that this is e^z with the first term ommited
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I Closed My Eyes
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Report Thread starter 4 weeks ago
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(Original post by DFranklin)
Surely this is just the series for e^z with the first term omitted (so = exp(z) - 1), so you're done?

Have to say, seems pretty advanced for year 9...
(Original post by RDKGames)
To add to the above, note that \dfrac{z^k}{k!} \cdot \dfrac{z}{k+1} = \dfrac{z^{k+1}}{(k+1)!} from which it should be obvious that this is e^z with the first term ommited
Ahhhhh thanks, I've just realised I was being quite silly and completely missed that

Also realised I accidentally put yr9 tag on haha
Last edited by I Closed My Eyes; 4 weeks ago
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