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.



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.


$$F(x)=\int_0^x f(y)dy=\sum_{t=1}^n\int_0^xty^{t-1}dy=\sum_{t=1}^nx^t=\textrm{you know how ...}$$

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


We recall the power rule, which states:

$$\frac{d}{dx} \left[x^n\right] = nx^{n-1}$$


$$\frac{d}{dx}\left[\sum_{t=1}^n x^t\right] = \sum_{t=1}^n tx^{t-1}$$

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:


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

  • $\begingroup$ 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 $0.01S$. $\endgroup$ – user62487108 Jan 15 at 2:27
  • 1
    $\begingroup$ @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 $\endgroup$ – Henry Jan 15 at 9:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.