Is there a closed form of the series
$$ \sum_{d=0}^D F^d $$
where D is a finite integer, not $\infty$, like in power series?
Asked
Active
Viewed 32 times
0
-
1It is a high-school formula: $;\dfrac{1-F^{D+1}}{1-F}$. – Bernard Mar 31 '20 at 20:20
-
1Is $F$ a number or something else? – Mnifldz Mar 31 '20 at 20:25
-
@Mnifldz: Yes, it is an integer. – Make42 Mar 31 '20 at 21:30
-
@Bernard: It is? Sorry, I couldn't find it and it is not obvious to me to be honest. Does it have a name, so I can search for the proof? – Make42 Mar 31 '20 at 21:32
-
It is directy derived from the factorisation formula $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1}) with $a=1$. I don't whether it has a worldwide name. I learnt it at the end of middle-school, and it is user later for the sum of a geometric series. – Bernard Mar 31 '20 at 22:12
1 Answers
1
The method for demonstrating the formula given by Bernard in the comments above is called telescoping. The trick is to observe that the sum can be written as:
$$ S \;\; =\;\; \sum_{d=0}^D F^d \;\; =\;\; 1 + F + F^2 + \ldots + F^D. $$
If I multiply $S$ by an additional copy of $F$ we obtain:
$$ FS \;\; =\;\; F\left (1 + F + F^2 + \ldots + F^D \right ) \;\; =\;\; F + F^2 + \ldots + F^D + F^{D+1}. $$
We can now subtract one equation from the other to obtain the telescoping effect: \begin{eqnarray*} S - FS & = & \left (1 + F + F^2 + \ldots + F^D \right ) - \left( F + F^2 + \ldots + F^D + F^{D+1} \right ) \\ & = & 1 - F^{D+1}. \end{eqnarray*}
Notice that the above subtraction eliminates all but the first and last terms. Acknowledging that $S$ was the sum we obtained, we can isolate it by noticing that $S - FS = S(1-F)$. Dividing both sides we find:
$$ S \;\; =\;\; \sum_{d=0}^D F^d \;\; =\;\; \frac{1 - F^{D+1}}{1-F}. $$
Mnifldz
- 12,880