# 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.

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 $$0.01\cdot f(0.99)=0.01\cdot F'(0.99)$$

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.

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: $$S= 100 -(100+n) \cdot (0.99)^{n}$$

• Sorry, am a little lost here. I understood everything up until line 3. What do you mean by subtraction? I can't really see how you got $0.01S$. – user62487108 Jan 15 at 2:27
• @user62487108 Subtracting the second line from the first you get $$S - 0.99S = 0.01 \cdot \sum\limits^{n-1}_{t=0}(t+1) \cdot (0.99)^{t} - 0.01 \cdot \sum\limits^{n}_{t=1}t \cdot (0.99)^{t}$$ which gives the third line with $0.01S$ on the left hand side – Henry Jan 15 at 9:18