Closed form for the n-th term I was wondering... Is there any closed form for the partial sums of this serie?
$$\sum_{k=1}^{n-1} k \frac{1}{n-1}$$

Notice, $$\sum_{k=1}^{n-1}k\frac{1}{n-1}$$
$$=\frac{1}{n-1}\sum_{k=1}^{n-1}k$$
This A.P. $\ \sum_{k=1}^{n-1}k$ can be further simplified as follows
$$=\frac{1}{n-1}(1+2+3+\ldots +(n-1))$$
$$=\frac{1}{n-1}\left(\frac{(n-1)}{2}(n-1+1)\right)$$
$$=\frac{1}{n-1}\left(\frac{n(n-1)}{2}\right)$$
$$=\color{red}{\frac{n}{2}}$$
