Closed form expression for $\sum_{k=0}^{n} \alpha^{-k}$?

I have seen this sum a couple of times now in the textbook and in the problems book but I don't know how to solve it. I know this might be more suited for math.stackexchange but since this is a fairly common sum and is related I'm pretty sure someone can help. Thank you!

• I'm not sure I undestand what you need help with. The equation in the red box is equal to $\alpha^n(1+1/\alpha+1/\alpha^2 + \ldots + 1/\alpha^n)$. Is that helpful? – MBaz Oct 16 '17 at 18:58

\begin{align} (1-x)\sum_{k=0}^{n}x^k &= \sum_{k=0}^{n}x^k - x\sum_{k=0}^{n}x^k \\ \\ &= \sum_{k=0}^{n}x^k - \sum_{k=1}^{n+1}x^k \\ \\ &=1 + x+x^2+ \dots+x^n \ \\ & \qquad -x-x^2-\dots-x^n-x^{n+1}\\ \\ &= 1 - x^{n+1} \end{align}

Thus if $x \ne 1$, dividing both sides by $(1-x)$ results in $$\sum_{k=0}^{n}x^k = \frac{1-x^{n+1}}{1-x}$$

Replace $x = \alpha^{-1}$, it becomes $$\sum_{k=0}^{n}\alpha^{-k} = \frac{1-(1/\alpha)^{n+1}}{1-1/\alpha}$$

• it seems to me that this answer suffices to answer the question. (but this summation is in the calculus textbook. any college sophomore should know it.) it should get the check mark. – robert bristow-johnson Oct 17 '17 at 0:10

@AlexTP provides the proof for the summation from $k=0$ to $k=n$. Let me put the more general case: Assume that the integers $a,b$ are finite, then for any $\beta$ we have $$\sum_{k=a}^{b} \beta^k = \frac{ ~~\beta^a - \beta^{b+1} }{ 1 - \beta}$$

When any of the summation limits $a$ or $b$ are infinite, then a limiting process should be considered and we have: $$\sum_{k=a}^{\infty} \beta^k = \frac{ ~~\beta^a }{ 1 - \beta}$$ for $|\beta| < 1$ , or else it won't converge. And for the infinity at the lower limit we have:

$$\sum_{k=-\infty}^{b} \beta^k = \frac{ - \beta^{b+1} }{ 1 - \beta}$$ for $|\beta| > 1$ , or else it won't converge.

Applying to your case with $a=0$, $b=n$ and $\beta = \alpha^{-1}$ :

$$\alpha^{n} \sum_{k=0}^{n} (\alpha^{-1})^k = \alpha^{n} \frac{ 1 - \alpha^{-n-1} }{1 - \alpha^{-1} } = \frac{ \alpha^{n+1} - 1 }{\alpha - 1 }= \frac{ 1 - \alpha^{n+1} }{1 - \alpha }$$

• Does the first summation apply for $\beta=1$? It looks like it should be equal to $b-a+1$, assuming $b \geq a$, but that's not obvious from the RHS. – MBaz Oct 16 '17 at 21:31
• neither of them shall be used for $\beta=1$ . However for the first sum you can get the correct result of course using the limit approach as follows: $$\sum_{k=a}^{b} 1^k = \lim_{x \to 1} \frac{x^a - x^{b+1}}{1-x}$$ Evaluated using L'hopital approach as: $$\sum_{k=a}^{b} 1 = \lim_{x \to 1} \frac{a x^{a-1} - (b+1)x^b}{-1} = b+1-a$$ – Fat32 Oct 16 '17 at 22:12
• Thanks. The reason for my comment is that you state that the summation formula is valid for "any $\beta$". – MBaz Oct 16 '17 at 22:21
• :-D Sorry if I came across the wrong way... I wanted to point out a way to improve your answer, and maybe get you to work out the details without me having to ;) – MBaz Oct 16 '17 at 23:06
• It is silly to work out the value of $$\sum_{k=a}^b \beta^k$$ when $\beta = 1$ by starting from the fancy formula and using L'Hopital's rule etc. More directly, $$\sum_{k=a}^b \beta^k\big|_{\beta = 1}=\sum_{k=a}^b 1^k = \sum_{k=a}^b 1 = b-a+1,$$ no muss, no fuss. – Dilip Sarwate Oct 17 '17 at 2:45