Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to prove this equation. $$ \sum_{k=0}^{i-2} \left( e \space α(k+1)\space\frac{(-1)^{i+k+2}}{(i-k-2)!} \right) = \sum_{k=0}^{i-2} \frac{(i-k)^k}{k!} \space e^{i-k} (-1)^k\space where,\space α(i) = \sum_{k=0}^{i-1}\left(\frac{(i-k)^k}{k!}e^{i-k}(-1)^k\right) \space\space\space\space (1)$$

This equation holds. I have confirmed it by using a math tool, but I don't know how the left side can be transformed into the right side of the equation. I need some formulas to prove this equation.

The left side of the equation includes series of series $\sum_{k=0}^{i-2} e \space α(k+1) $, so it is very difficult for me to transform the series of series into the series in the right side.

The equation can be rewritten as follows,

$$ \sum_{k=0}^{i-2} \left( e \space α(k+1)\space\frac{(-1)^{i+k+2}}{(i-k-2)!} \right) = α(i) - \frac{(-1)^{i+1} \space e}{(i-1)!} \space\space\space(2)$$ $$ α(i) = \sum_{k=0}^{i-2} \left( e \space α(k+1)\space\frac{(-1)^{i+k+2}}{(i-k-2)!} \right) + \frac{(-1)^{i+1} \space e}{(i-1)!} \space\space\space(3)$$

Any hint that will lead me to the correct proof will be highly appreciated.

For more information, this $α(i)$ function came from my previous question.

share|improve this question
Since the lhs is a polynomial of $e$ (which is transcendental) the coefficients of each degree $e^k$ should be equal. –  Andrew Jul 16 '12 at 5:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.