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Consider $N_1 = \sum_{k=1}^K kr^k$ and $N_2 = \sum_{k=0}^K kr^k$ (assuming $r \neq 1$).

These two equations are equivalent since the first term in $N_2$ will always be zero due to the multiplication by $k$.

However, solving now:

$N_1 = \sum_{k=1}^K kr^k$

$N_1 = r\sum_{k=1}^K kr^{k-1}$

$N_1 = r\left(\sum_{k=1}^K r^{k}\right)^{'}$ (derivate trick)

Using geometric progression standard form:

$N_1 = r\left(r\frac{1-r^{K}}{1-r}\right)^{'}= r\left(\frac{r-r^{K+1}}{1-r}\right)^{'}$

Now solving for $N_2$:

$N_2 = \sum_{k=0}^K kr^k$

$N_2 = r\sum_{k=0}^K kr^{k-1}$

$N_2 = r\left(\sum_{k=0}^K r^{k}\right)^{'}$ (derivate trick)

Using geometric progression standard form:

$N_2 = r\left(\frac{1-r^{K+1}}{1-r}\right)^{'}$

Clearly, $N_1 \neq N_2$ based on this expansion. Where is the flaw in the logic?

Allan_ZA
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2 Answers2

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Your formula for a geometric sum is not correct. The correct formulas are $\sum\limits_{k=0}^{n} r^{k}=\frac {1-r^{n+1}} {1-r}$ and $\sum\limits_{k=1}^{n} r^{k}=\frac {r-r^{n+1}} {1-r}$.

Kavi Rama Murthy
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As observed in another answer, you have written $\newcommand{ddr}{\frac{\mathrm d}{\mathrm dr}}r^{K-1}$ where you should have written $r^{K+1}$ in the last equation.

That might just be a transcription error. The formulas might still look like different results at first glance.

But look what happens if you put the correct formula in place and examine it more carefully. What actually is the difference between the two formulas whose derivatives you are taking?

Taking $K$ as a constant and $r \in (-1,1)$ (as we need to do in order for your various manipulations of the geometric series to be valid),

$$ \frac{1-r^{K-1}}{1-r} - \frac{r-r^{K-1}}{1-r} = \frac{1-r}{1-r} = 1, $$

and therefore

$$ \ddr\left(\frac{1-r^{K-1}}{1-r}\right) - \ddr\left(\frac{r-r^{K-1}}{1-r}\right) = \ddr\left(\frac{r-r^{K-1}}{1-r} - \frac{r-r^{K-1}}{1-r}\right) = 0. $$

So it turns out that

$$ N_2 = r \ddr\left(\frac{1-r^{K-1}}{1-r}\right) = r \ddr\left(\frac{r-r^{K-1}}{1-r}\right) = N_1. $$

David K
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