How can I evaluate $0.01 \cdot \sum^{n}_{t=1}t \cdot (0.99)^{t-1}$? How can I evaluate the summation $$0.01 \cdot \sum^{n}_{t=1}t \cdot (0.99)^{t-1}$$
I am aware that this is almost in the form of a geometric summation, but I am unsure how to proceed and what to do. Any help would be greatly appreciated.
 A: Hint:
Define the function $f(y)=\sum_{t=1}^n ty^{t-1}$, $y\in[0,1]$
Integrate on $[0,x]$ term by term, take the sum, then differentiate it.

Notes.
$$F(x)=\int_0^x f(y)dy=\sum_{t=1}^n\int_0^xty^{t-1}dy=\sum_{t=1}^nx^t=\textrm{you know how ...}$$
Then
$$
0.01\cdot f(0.99)=0.01\cdot F'(0.99)
$$
A: We recall the power rule, which states:
$$\frac{d}{dx} \left[x^n\right] = nx^{n-1}$$
Hence:
$$\frac{d}{dx}\left[\sum_{t=1}^n x^t\right] = \sum_{t=1}^n tx^{t-1}$$
Recalling that $\sum_{t=1}^{n} x^t = (x-x^{n+1})/(1-x)$ (see if you can derive this yourself! Hint: multiply both sides by $(1-x)$) to get your desired expression we solve for:
$$0.01\left(\frac{d}{dx}\left[\frac{x-x^{n+1}}{1-x}\right]\right)$$
evaluated at $x=0.99$.
The first time you see term-by-term differentiation or integration to manipulate a series it seems like such an illegal, out-there move, but it's actually a regular tool in the toolbox. Once you see it once, you see it everywhere.
A: If $S=0.01 \cdot  \sum\limits^{n}_{t=1}t \cdot (0.99)^{t-1}=0.01 \cdot  \sum\limits^{n-1}_{t=0}(t+1) \cdot (0.99)^{t}$
then $0.99S=0.01 \cdot \sum\limits^{n}_{t=1}t \cdot (0.99)^{t} $
so by subtraction $0.01S = 0.01 + 0.01 \cdot \sum\limits^{n-1}_{t=1}  (0.99)^{t} - 0.01 \cdot n \cdot (0.99)^{n}$
i.e. $S = 1  -  n \cdot (0.99)^{n} +  \sum\limits^{n-1}_{t=1}  (0.99)^{t}$
and then $0.99S=0.99  - 0.99 \cdot  n \cdot (0.99)^{n} +  \sum\limits^{n}_{t=2}  (0.99)^{t}$
so by subtraction $0.01S = 0.01 - 0.01  \cdot n\cdot(0.99)^{n} +  0.99 -  (0.99)^{n}$
i.e. $S=1-n\cdot(0.99)^{n}+ 100\cdot  0.99 -100\cdot  (0.99)^{n}$
or more tidily:  $$S= 100 -(100+n) \cdot (0.99)^{n}$$
