How to evaluate the following sum? $\displaystyle \frac{100}{100}\cdot 1 + \frac{100\cdot 99}{100^2}\cdot 2 + \frac{100\cdot 99\cdot 98}{100^3}\cdot 3 + \frac{100\cdot 99\cdot 98\cdot 97}{100^4}\cdot 4 + \ldots + \frac{100\cdot 99\cdot 98\cdot  \ldots\cdot 1}{100^{100}} $
What I have tried:
Above sum can be written as $\displaystyle S =  \sum \limits_{i=1}^{100} \frac{100\cdot 99\cdot  ... \cdot (100-i+1)}{100^i} \cdot  i$
$\displaystyle =\sum \limits_{i=1}^{100} \frac{^{100}P_i}{100^i} \cdot i $
I am completely stuck here. If not for the $^{100}P_i$ term, this would have been a simple arithmetico-geometric series. How should I proceed?
 A: The sum can be written as
$$\sum_{i=0}^{100}\frac{100!i}{(100-i)!\cdot100^i}$$
$$=100!\sum_{i=0}^{100}\frac{100-i}{i!\cdot100^{100-i}}$$
$$=\frac{100!}{100^{100}}\sum_{i=0}^{100}\frac{(100-i)\cdot100^i}{i!}$$
$$=\frac{100!}{100^{100}}\sum_{i=0}^{100}\left(\frac{100^{i+1}}{i!}-\frac{100^i}{(i-1)!}\right)$$
which telescopes. Going to the  second line from the first, I've substituted $100-i$ for $i$.
Edit in response to OP's comments
The substitution does not, in fact, change anything else. The notation $\sum_{i=0}^{100}f(i)$ simply tells us to sum the terms over the natural numbers ranging from $0 $ to $100$. When you consider $\sum_{i=0}^{100}f(100-i)$, you can see that it gives you the same set of terms, (i.e. $f(100),f(99),\cdots,f(0)$). And the sigma notation is used as "counting up", so we do not change the limits.
This is indeed neither an arithmetic series nor a geometric series. Are you familiar with telescoping sums? Notice that the expression inside the summation in the final line has the form $f(i+1)-f(i)$. Let us define $$f(i)=\frac{100^i}{(i-1)!}$$
So we have to evaluate $$\sum_{i=0}^{100}\left(f(i+1)-f(i)\right)$$
$$=f(1)-f(0)+f(2)-f(1)+f(3)-f(2)+\cdots+f(100)-f(99)+f(101)-f(100)$$
After cancelling, we have left $f(101)-f(0)$. Now recall $$f(i)=\frac{100^i}{(i-1)!}=\frac{100^i\cdot i}{i!}$$
Then $f(101)=\dfrac{100^{101}\cdot101}{101!}$ and $f(0)=0$.
Thus, the sum is $$\frac{100!}{100^{100}}\times\frac{100^{101}\cdot101}{101!}$$
$$=\boxed{100}$$
Also note that we can change from $\sum_1^{100}$ to $\sum_0^{100}$ because the term corresponding to $i=0$ is just $0$ and does not change the sum.
