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I want to prove $\displaystyle\sum_{n=k}^{+\infty} \left[\begin{array}{c}n\\k\end{array}\right]\frac{z^n}{n!}=\frac{\log^k(1/(1-z))}{k!}$ by induction.

For $k=0$ or $k=1$ it works, suppose it works for some $k$, the question is if $\displaystyle\sum_{n=k+1}^{+\infty} \left[\begin{array}{c}n\\k+1\end{array}\right]\frac{z^n}{n!}=\frac{\log^{k+1}(1/(1-z))}{(k+1)!}$, so:

$\displaystyle\frac{\log^{k+1}(1/(1-z))}{(k+1)!}=\frac{1}{k+1}\cdot\frac{\log^k(1/(1-z))}{k!}\cdot\log(1/(1-z))= \\ \displaystyle\frac{1}{k+1}\cdot \left( \sum_{n=k}^{+\infty} \left[\begin{array}{c}n\\k\end{array}\right]\frac{z^n}{n!} \right)\cdot \left( \sum_{n=k}^{+\infty} (n-1)! \frac{z^n}{n!} \right)$ and if I'm not wrong we have convolution of exponential generating functions and it's equal to: $\displaystyle\frac{1}{k+1}\cdot \sum_n\sum_i{n \choose i}\left[\begin{array}{c}i\\k\end{array}\right](n-i+1)!\frac{z^n}{n!}$ but I don't know what to do next.. or maybe I went wrong way?

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    could anyone write how to proof this using generating functions?2012-03-21

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Ok, let's try using induction as requested (I have a feeling it would be easier to use generating functions).

Assume (for $|z| <1$) $S_k(z)=\sum_{n=k}^{\infty} \left[\begin{array}{c}n\\k\end{array}\right]\frac{z^n}{n!}=\frac{\log^k[1/(1-z)]}{k!}.$

We like to evaluate $S_{k+1}(z)=\sum_{n=k+1}^{\infty} \left[\begin{array}{c}n\\k+1\end{array}\right]\frac{z^n}{n!}.$ Note the recurrence relation $\begin{bmatrix} n \\ k\end{bmatrix} = \begin{bmatrix} n+1 \\ k+1\end{bmatrix} - n \begin{bmatrix} n \\ k+1\end{bmatrix}. $ With that we have $\begin{align} S_k(z) &= \sum_{n=k}^{\infty}\begin{bmatrix} n+1 \\ k+1\end{bmatrix} \frac{z^n}{n!} - \sum_{n=k}^\infty n \begin{bmatrix} n \\ k+1\end{bmatrix} \frac{z^n}{n!}\\ &= \sum_{n=k+1}^\infty \begin{bmatrix} n \\ k+1\end{bmatrix} \frac{1}{(n-1)!} ( z^{n-1} - z^n) \\ &= (1- z)\frac{d}{d z} S_{k+1}(z), \end{align}$ where we used the fact that $\begin{bmatrix} k \\ k+1 \end{bmatrix} =0$.

Solving for $S(k+1)$, we have (employing the substitution t=\log[1/(1-z')] and the trivial fact that $S_{k+1}(0) =0$) \begin{array} \\ S_{k+1}(z) &= \int_0^z \!dz'\, \frac{S_k(z')}{1-z'} = \int_0^z\! dz'\frac{\log^k[1/(1-z')]}{k! (1-z')} \\ &= \int_0^{\log[1/(1-z)]} \!dt\, \frac{t^k}{k!} = \frac{\log^{k+1} [1/(1-z)]}{(k+1)!}.\end{array}

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    You can just put in the formula which you know for $S_{k+1}$ into the expression $S_k=(1+z)\partial_z S_{k+1}$ and see that this relation is true. Together with $S_{k+1}(0)=S_k(0)=0$ this concludes the proof.2012-03-20