I was thinking about the wheat and chessboard story and thinking of the total number of grains of wheat…

$$\sum_{n=0}^{63} 2^n$$

And wondered if there is a name for a sum like this?

$$2^0 + 2^1 +2^2 + 2^3 + \cdots$$

  • 26
    $\begingroup$ Geometric? ${}{}$ $\endgroup$
    – user251257
    Aug 14 '16 at 14:35
  • $\begingroup$ The first on is a finite geometric series. The second is an infinite geometric series. To call it "geometric" means there is a common ratio, i.e. the number you multiply each term by, to get the next term, is the same in every case. $\qquad$ $\endgroup$ Aug 14 '16 at 17:51
  • $\begingroup$ If infinite, the second one is not convergent -- it diverges quite badly. @MichaelHardy $\endgroup$
    – Clement C.
    Aug 14 '16 at 19:23
  • 2
    $\begingroup$ Whether it diverges badly or well seems a matter of opinion: I'd go for the latter. It diverges really quite well indeed. $\endgroup$ Aug 14 '16 at 19:59
  • 3
    $\begingroup$ @Dewi, if adjectives are a matter of opinion, I'd pick "spectacularly" over the two other choices you mentioned. $\endgroup$ Aug 14 '16 at 22:48

This is called a geometric sum/series according as $N$ is finite or not. More generally, it is of the form $$ \sum_{n=0}^Nar^n$$ where $N$ may be "equal to" infinity. In the case where $N$ is infinite, if the series is to converge, we require $\lvert r\rvert<1$. Furthermore, for $N$ finite $$ \sum_{n=0}^Nar^n=a\times\frac{1-r^{N+1}}{1-r},\:\:\:\: r\ne1$$ and $$ \sum_{n=0}^\infty ar^n=\frac{a}{1-r},\:\:\:\:\lvert r\vert<1.$$

  • 3
    $\begingroup$ This is wrong. For the finite case, the sum shall be $a\frac{1-r^{N+1}}{1-r}$ since there are $N+1$ terms. $\endgroup$
    – Kun
    May 4 '19 at 21:02

It's mainly a well known value $2^{n+1}-1$ . Write it in base $2$.

It is also OEIS sequence A000225 :

$2^n - 1.$ (Sometimes called Mersenne numbers, although that name is usually reserved for A001348.)

  • 10
    $\begingroup$ This answers the specific case of $r=2$, but not the general case asked in the title. The sum of increasing powers is a geometric sum -- the specific name of Mersenne numbers is coincidental in this question, and giving the closed form without the explanation (first part of this answer) sheds basically no light whatsoever on the why. $\endgroup$
    – Clement C.
    Aug 14 '16 at 15:08
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    $\begingroup$ The OP asked if there "was a name for a sum like that." (Emphasis mine) Your answer gives a value without explanation, and a name for this exact sum which is not generalizable to any other ratio. As such, I feel it fails to answer the question. $\endgroup$
    – Clement C.
    Aug 14 '16 at 16:19
  • $\begingroup$ Indeed, it's my opinion - this is why I wrote a comment to explain it. $\endgroup$
    – Clement C.
    Aug 14 '16 at 16:33

As you are looking for novel names, this is also a simple case of the more generic notion of hypergeometric series $$\sum_{k}r_k \,,$$ with $r_0 = 1$, and the ratio of two consecutive terms is a rational function, a ratio of two polynomials $P$ and $Q$ in the summation index $k$

$$ \frac{r_{k+1}}{r_k}= \frac{P(k)}{Q(k)}\,. $$

In your case, you can choose $P$ and $Q$ such that their ratio is equal to $2$. When the ratio is constant, it is called a geometric series (as answered here). As a reminder, it is a sum of terms in geometric progression (se.math) like $1,r,r^2,r^3,\ldots$, whose name (the geometry part) is illustrated by the following figure:

geometric progression

Hypergeometric series are also connected to chess. A rook is a move on a chessboard. Some have developed studies some types of permutations as the placement of a number of rooks on a chessboard-like grid, see for instance Rook theory and hypergeometric series, J. Haglund, 1996.


The formula for the sum of the first $N+1$ terms can be derived as following:


$$S(N)=\sum_{n=0}^{N} r^n$$

Then distributing an $r$ (which is independent of $n$) along the sum we have,

$$rS(N)=\sum_{n=0}^{N} r^{n+1}$$

Note (1):

$$S(N+1)=\sum_{n=0}^{N+1} r^n=r^0+\sum_{n=0}^{N} r^{n+1}$$ $$=1+rS(N)$$

Also note,

$$S(N+1)-S(N)=\sum_{n=0}^{N+1} r^n-\sum_{n=0}^{N} r^n=r^{N+1}$$

As all terms but $r^{N+1}$ will reduce each other out to zero.



Hence we have from (1) and from the above:






Now we multiply by a special form of $1$ to get:

$$S(N)=\sum_{n=0}^{N} r^n=\left( \frac{-1}{-1} \right) \left( \frac{r^{N+1}-1}{r-1} \right)=\frac{1-r^{N+1}}{1-r}$$

And hence (distributing out an $a$ which is independent of $n$) we get the formula for the geometric sum,

$$\sum_{n=0}^{N} ar^n=a\frac{1-r^{N+1}}{1-r}$$

Note though our formula only works for $r \neq 1$ (our manipulations are okay until we get to dividing by $0=1-r=1-1$ for $r=1$).


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