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I have to calculate this \begin{align} & (1-p)^{N-1}+\frac{1}{2}C_{N-1}^{N-2}(1-p)^{N-2}p+\frac{1}{3}C_{N-1}^{N-3}(1-p)^{N-3}p^{2}+...+\frac{1}{N-2}C_{N-1}^{1}(1-p)p^{N-2}\\ & +\frac{1}{N}p^{N-1} \end{align} where $C_{n}^{m}=\frac{n!}{m!(n-m)!}$.

Can anyone help to sum this up? Thank you.

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    Coefficient of second last term should read $\frac 1{N-1}$ instead of $\frac 1{N-2}$. – 2017-02-27

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The given sum can be written as:

\begin{align*} \sum_{k=1}^{N} \frac 1k \binom{N-1}{N-k} p^{k-1}(1-p)^{N-k} = (1-p)^{N-1} \sum_{k=1}^{N} \frac 1k \binom{N-1}{N-k} \left(\frac{p}{1-p}\right)^{k-1} \end{align*}

Set $r=\frac{p}{1-p}$ and observe that

\begin{align*} \sum_{k=1}^{N} \frac 1k \binom{N-1}{N-k} r^k &= \sum_{k=1}^{N} \binom{N-1}{N-k} \int_0^r x^{k-1} \mathrm{d}x \\ &= \int_0^r \sum_{k=1}^{N} \binom{N-1}{N-k} x^{k-1} \mathrm{d}x \\ &= \int_0^r \sum_{k=1}^{N} \binom{N-1}{k-1} x^{k-1} \mathrm{d}x \\ &= \int_0^r \sum_{k=0}^{N-1} \binom{N-1}{k} x^{k} \mathrm{d}x \\ &= \int_0^r (1+x)^{N-1} \mathrm{d}x \\ &= \frac{(1+r)^N - 1}{N} \end{align*}

Thus, the required sum is $$(1-p)^{N-1} \left(\frac{1-p}{p}\right) \frac{\left(\frac{1}{1-p}\right)^N - 1}{N} = \frac{1-(1-p)^N}{pN}$$

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    thank you. I have worked out another way. use $c_{N-1}^{N-1-k}=c_{N-1}^{k}$ and $\frac{1}{k+1}c_{N-1}^{k}=\frac{1}{N}c_{N}^{k+1}$. Then we can use $(1-p+p)^{N}=\sum_{k=0}^{N}p^{k}(1-p)^{N-k}$. – 2017-02-27
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    Oh yeah! That's another slick way. Pleasure was all mine. – 2017-02-27
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    From $\int_0^r \sum_{k=1}^{N-1} \binom {N-1}{k-1}x^kdx$ to $\int_0^r \sum_{k=0}^{N-1} \binom {N-1}kx^kdx$, you seem to have added a term - $\int_0^r \binom {N-1}{N-1}x^{N-1}dx$ – 2017-02-27
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    @AnotherJohnDoe Thanks for pointing that out. I have made the corrections. – 2017-02-27
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    @congmingniao You can accept this solution if you're satisfied, by checking the tick mark right below the upvote/downvote button for this answer. – 2017-02-27
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Note that $$\frac 1{r+1}\binom {N-1}{N-1-r}=\frac 1{r+1}\binom {N-1}r=\frac 1N\binom N{r+1}$$ and $$(1-p)^{N-1-r}\;p^r=\frac {(1-p)^N}p\left(\frac p{1-p}\right)^{r+1}$$ Hence $$\begin{align} \sum_{r=0}^{N-1}\frac 1{r+1}\binom {N-1}{N-1-r}(1-p)^{N-1-r}p^r &=\sum_{r=0}^{N-1}\frac 1N\binom N{r+1}\frac {(1-p)^N}p\left(\frac p{1-p}\right)^{r+1}\\ &=\frac {(1-p)^N}{Np}\sum_{r=0}^{N-1}\binom N{r+1}\left(\frac p{1-p}\right)^{r+1}\\ &=\frac {(1-p)^N}{Np}\sum_{r=1}^{N}\binom N{r}\left(\frac p{1-p}\right)^{r}\\ &=\frac {(1-p)^N}{Np}\cdot \left[\left(1+\frac p{1-p}\right)^N-1\right]\\ &=\color{red}{\frac {1-(1-p)^N}{Np}} \end{align}$$