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Please, I need to know the proof that

$$\left(\sum_{k=0}^{\infty }\frac{n^{k+1}}{k+1}\frac{x^k}{k!}\right)\left(\sum_{\ell=0}^{\infty }B_\ell\frac{x^\ell}{\ell!}\right)=\sum_{k=0}^{\infty }\left(\sum_{i=0}^{k}\frac{1}{k+1-i}\binom{k}{i}B_in^{k+1-i}\right)\frac{x^k}{k!}$$

where $B_\ell$, $B_i$ are Bernoulli numbers.

Maybe we should replace $k$ with $j$?

Anyway, I need to prove how to move from the left to right.

Thanks for all help.

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    $\begingroup$ planet math - Math for the people, by the people. and People are never wrong :P $\endgroup$ – Mula Ko Saag May 18 '13 at 17:12
  • $\begingroup$ sry but i dont know what happen $\endgroup$ – mhd.math May 18 '13 at 17:12
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$$\left(\sum_{k=0}^{\infty} \dfrac{n^{k+1}}{k+1} \dfrac{x^k}{k!} \right) \left(\sum_{l=0}^{\infty} B_l \dfrac{x^l}{l!}\right) = \sum_{k,l} \dfrac{n^{k+1}}{k+1} \dfrac{B_l}{k! l!} x^{k+l}$$

$$\sum_{k,l} \dfrac{n^{k+1}}{k+1} \dfrac{B_l}{k! l!} x^{k+l} = \sum_{m=0}^{\infty} \sum_{l=0}^{m} \dfrac{n^{m-l+1}}{m-l+1} \dfrac{B_l}{(m-l)! l!} x^{m}$$ This gives us $$\sum_{m=0}^{\infty} \sum_{l=0}^{m} \dfrac{B_l}{(m-l+1)! l!}n^{m-l+1} x^{m} = \sum_{m=0}^{\infty} \left(\sum_{l=0}^m \dfrac1{m-l+1} \dbinom{m}{l} B_l n^{m-l+1}\right)\dfrac{x^m}{m!}$$

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The right-hand side is just the Cauchy product of the two series on the left-hand side. Generally, the Cauchy product of $\sum_{k=0}^\infty a_k$ and $\sum_{l}^\infty b_l$ is $$\sum_{j=0}^\infty\sum_{i=0}^j b_i a_{j-i}.$$ Now you should have no problem finishing the problem. (Change $j$ to $k$ at the end to match your right hand side.)

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