Series where the sum is 0

Question

I'm not a mathematician and don't know a lot about series but I'm trying to figure out if there is an infinite geometric series that starts at 1 (or any other positive number), has fractions always subtracted from it, and has a sum of 0.

Answer 1

Because MSE would prefer not to have questions answered solely in the comments, it is recommended to transfer the most helpful comments to a bona-fide answer. I can make this answer CW (community-wiki) upon request.

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A comment by [@Ethan Bolker] said that $1-\frac 12 - \frac 14 - \frac 18 - \ldots$ is an infinite series that converges to $0$.

A comment by [@M W] remarked that more generally, for any infinite series $\sum_{i=0}^\infty b_i$ which converges to $L$, the infinite series $L-b_0-b_1-b_2-\ldots$ converges to $0$.

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Since the OP asked in the comments about this: if we have an infinite sum $\sum_{i=0}^\infty b_i$, the phrases:

• the infinite series $\sum_{i=0}^\infty b_i$ converges to/approaches $L$
• the infinite sum $\sum_{i=0}^\infty b_i$ equals/evaluates to $L$
• the partial sums $b_0, b_0+b_1, b_0+b_1+b_2, \ldots, \sum_{i=0}^n b_i, \ldots$ converge to/approach $L$

ALL mean the same thing, and can be interchangeably used. Usually, if we write an expression which ends with "dot dot dot", e.g. $1+\frac 12+\frac 14 + \ldots$, we typically would say that "converges to/approaches" the limit $L$ instead of it "equals" $L$, but saying "equals" is also fine (I wouldn't say it's wrong, but "converges to" just sounds a bit better).

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I maybe can also add a picture to this discussion. In the below picture, we have purple squares of area $(\frac 12)^2, (\frac 14)^2, \ldots$ (in decreasing order). Because there are a total of 3 squares of the same dimensions (a purple square, with a yellow squares directly above and to the right of it, all of the same dimensions), the total sum of the purple areas is $L:=\frac 13$.

So, you can imagine that we have the purple region with area $L=\frac 13$, and we start taking away squares one by one (taking the biggest one first, and so on). Then eventually taking away all the purple squares we get $\frac 13- (\frac 12)^2 - (\frac 14)^2- \ldots$ converges to $0$.