# Infinite series with finite sum [duplicate]

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Assume that I have 1 unit of something and then I add ½ unit resulting in a total of 1.5 units. Then I add half of the half (0.25 units) for a total of 1.75 units. Then I add the half of the half of the half (0.125 units) resulting in 1.875 units. IIUC, the result will be finite and have a limit but what is it and is there a general formula I'm looking for?

The problem seems similar to a Riemann sum where an infinite sequence has a finite value, or similar to Xeno's paradox about Achilles and the tortoise.

## marked as duplicate by GNUSupporter 8964民主女神 地下教會, Ethan Bolker, Did sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 1 '18 at 15:11

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## 4 Answers

This is called a geometric series. In general, you have that

$$\sum\limits_{k=0}^\infty x^k = \frac{1}{1-x}$$ (Provided that $|x|<1$, otherwise the series does not converge)

For example $$1+\frac12+\frac14+... = \sum\limits_{k=0}^\infty \left(\frac12\right)^k = \frac{1}{1-1/2} = 2.$$

Your sum is a geometric series. This is visualized here: 1 + 1/2 + 1/4 + ...

Note that you are looking for

$$1 +\frac12 +\frac14+...=\sum_{k=0}^{\infty} \left(\frac12\right)^k$$

which is a geometric series and the sum for $|r|<1$ is given by $\sum\limits_{k=0}^\infty r^k = \frac{1}{1-r}$.

Note also that the proof is quite simple since

$$(1+r+r^2+...+r^n)(1-r)=1-r^{n+1}\implies \sum\limits_{k=0}^n r^k=\frac{1-r^{n+1}}{1-r}\to\frac1{1-r}$$

indeed the term $r^{n+1}$ vanishes for n large when $|r|<1$.

1) $S_m =$

$1+(1/2)^1+(1/2)^2+.....(1/2)^m.$

2) $(1/2)S_m=$

$\enspace \enspace \enspace \enspace (1/2)^1+(1/2)^2+.....(1/2)^m$

$+ .....(1/2)^{m+1}$.

Subtract :1)-2):

$(1/2)S_m= 1-(1/2)^{m+1}$, or

$S_m= 2- 2(1/2)^{m+1}.$

$\lim_{m \rightarrow \infty } 2(1/2)^{m+1}=$

$\lim_{m \rightarrow \infty} (1/2)^m= 0.$

Achilles racing against the turtle:

Since the turtle is slower it starts with an advantage .

1) They start racing:

Achilles catches up with the starting position of the turtle, the turtle has advanced a bit_1.

Achilles catches up with this bit _1, the turtle has advanced a bit_2.

Whenever Achilles catches up with the bit advanced, the turtle has advanced a bit more.

Conclusion: Speedy Achilles does not catch up with the turtle. (???)