How can I prove that $$ \sum_{i=0}^\infty \frac{z^{i+1}}{(i+1!)} = e^z - 1 $$
knowing that $$ \sum_{i=0}^\infty \frac{z^i}{i!} = e^z $$
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I dont see whats there to prove when you know that. – King Tut Feb 24 '18 at 16:58
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@KingTut, the two equations are different. The bottom one is only for additional information, not necessarily part of the proof. – RVC Feb 24 '18 at 17:01
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Given that liberty, why would anyone not use it. – King Tut Feb 24 '18 at 17:04
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Strategy: when you see sums written with $\sum$ write out the first few terms followed by an ellipsis ($\ldots$). That will often show you what's happening. – Ethan Bolker Feb 24 '18 at 17:17
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Thank you, @EthanBolker. Will try it out on paper now... – RVC Feb 24 '18 at 17:18
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Isn't $ \sum_{i=0}^\infty \frac{z^{i+1} } { (i+1)! } = \frac{z}{1} + \frac{z^2}{2!} + \frac{z^3}{3!}.... = e^z$?? – RVC Feb 24 '18 at 17:26
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1@RVC No, that sum is $e^z-1$. – Mark Viola Feb 24 '18 at 17:53
2 Answers
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Reindex: $$ \sum_{i=0}^\infty \frac{z^{i+1}}{(i+1)!}= \sum_{i=1}^\infty \frac{z^{i}}{i!}=\sum_{i=0}^\infty \frac{z^{i}}{i!}-1=e^z-1 $$
operatorerror
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@qbert, terribly sorry, but when you reindexed to i=1 to \infty, what happend to the i=0 term? or if you set the new index i=i+1, why didn't it start from -1? – RVC Feb 24 '18 at 17:14
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Hint:
$$\sum_{i=0}^{\infty}\frac{z^i}{i!}=1+\sum_{i=0}^{\infty}\frac{z^{i+1}}{(i+1)!}$$ because $$\sum_{i=0}^{\infty}\frac{z^i}{i!}=1+\frac{z^1}{1!}+\frac{z^2}{2!}+\dots=1+\left (\frac{z^1}{1!}+\frac{z^2}{2!}+\dots\right )=1+\sum_{i=0}^{\infty}\frac{z^{i+1}}{(i+1)!}$$
cansomeonehelpmeout
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Thank you, @cansomeonehelpmeout. But this is exactly my question, just rearranged... How did you take the 1 out of the sum? – RVC Feb 24 '18 at 17:17
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