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Is there a way to calculate this not term-by term ?

$$E_n =e^1+e^2+...+e^n$$

I tried developping :

$$E_n = \sum^\infty_{k=0}\frac{1^k}{k!}+\sum^\infty_{k=0}\frac{2^k}{k!}+...+\sum^\infty_{k=0}\frac{n^k}{k!}$$

It can be shortened to :

$$E_n = \sum^\infty_{k=0}\frac{1^k+2^k+...+n^k}{k!}$$

$$E_n = \sum^\infty_{k=0}\frac{\sum^n_{i=1}i^k}{k!}$$

Now, I don't know if there's a way to continue.

I searched around the internet and stumbled across the Faulhaber formula. Does this can help ?

Edit : ok i'm stupid I forgot about the geometric series :(

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  • $\begingroup$ I think there should be a formula to express the numerator in a more compact way $\endgroup$ Dec 25, 2016 at 15:06
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    $\begingroup$ Why not a geometric series? This is a common summation. $\endgroup$
    – Kaynex
    Dec 25, 2016 at 15:07
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    $\begingroup$ Response to edit: no, not stupid. Everyone misses easy stuff from time to time. $\endgroup$ Dec 25, 2016 at 15:19

2 Answers 2

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Each term is being multiplied by $e$, so this is called the partial sum of a geometric series, so the formula is: $$E_n=e\left(\frac{1-e^n}{1-e}\right)$$

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If:${a_n},{b_n}$ are sequences,and $a_{n+1}-a_n=A,\frac{b_{n+1}}{b_n}=B;A,B $ are constant and $B\neq 1$

Let:$S_n=\sum_\limits{k=1}^n{a_k\cdot b_k}$ ,then we have:$$B·S_n=\sum_\limits{k=1}^n a_k ·b_{k+1}$$ So:$$(1-B)·{S_n}=\sum_\limits{k=1}^n {(a_{k+1}-a_k)}b_{k+1}+a_1 b_1-a_n\cdot b_{n+1}$$ $$=A\sum_\limits{k=2}^n b_{k}+a_1 b_1-a_n\cdot b_{n+1}$$ $$=A\frac{b_2(1-B^n)}{1-B}+a_1 b_1-a_n\cdot b_{n+1}$$$$S_n=A\frac{b_2(1-B^n)}{(1-B)^2}+\frac{a_1 b_1-a_n\cdot b_{n+1}}{1-B}$$

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