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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.

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

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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.

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

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  • $\begingroup$ 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
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    $\begingroup$ @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

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