# Does i to the power of infinity =...0? **URL:** \ **Category:** Factual Questions\ **Tags:** science-math\ **Created:** [April 15, 2025, 8:49pm UTC](https://boards.straightdope.com/t/does-i-to-the-power-of-infinity-0/1017083 "2025-04-15T20:49:31Z")\ **Posts on this page:** 5\ **Page:** 2
**Author:** ![Civil\_Guy](https://avatars.discourse-cdn.com/v4/letter/c/d78d45/32.png) [@Civil\_Guy](https://boards.straightdope.com/u/Civil_Guy)\ **Post date:** [April 17, 2025, 1:50am UTC](https://boards.straightdope.com/t/does-i-to-the-power-of-infinity-0/1017083/21 "2025-04-17T01:50:25Z")
Infinity is simply not a single number; it’s more a process than anything else. My vote is ‘indeterminate’. ---
**Author:** ![EnolaStraight](https://avatars.discourse-cdn.com/v4/letter/e/41988e/32.png) [@EnolaStraight](https://boards.straightdope.com/u/EnolaStraight)\ **Post date:** [April 17, 2025, 2:06am UTC](https://boards.straightdope.com/t/does-i-to-the-power-of-infinity-0/1017083/22 "2025-04-17T02:06:07Z")
This was the point I was trying to articulate. As i to the power of x…x goes to infinity…you don’t just figure the “last number” of infinity gives you the answer. ALL the values of infinity…integers, fractions, decimals, irrational numbers…are taken as a whole. Which would generate ALL points on the Unit Circle. And _what point_ defines a circle? THE CENTER! At the origin…zero. The series 1+2+3+4+…=-1/12, if constructed as a parabola-like curve, has the vertex indeed at -1/12 [[1 + 2 + 3 + 4 + ⋯ - Wikipedia]](https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#Bibliography) Despite the series going out to infinity, -1/12 is a point which, arguably, is at the beginning of the curve. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [April 17, 2025, 2:52am UTC](https://boards.straightdope.com/t/does-i-to-the-power-of-infinity-0/1017083/23 "2025-04-17T02:52:18Z")
> [@Great\_Antibob](#): > > If you add an infinite set of terms together, we have a definition that calls it “convergent” if, basically and somewhat oversimplifying, we can come up with an answer most people would agree to. That’s not at all what “convergent” means, and a good thing, too, because most people would agree that adding up an infinite number of things, by any method, is nonsense. Convergence is something that people don’t know about until they’re taught it, and any of the other notions of infinite sums can also be taught. ---
**Author:** ![Great\_Antibob](https://avatars.discourse-cdn.com/v4/letter/g/e47c2d/32.png) [@Great\_Antibob](https://boards.straightdope.com/u/Great_Antibob)\ **Post date:** [April 17, 2025, 3:11am UTC](https://boards.straightdope.com/t/does-i-to-the-power-of-infinity-0/1017083/24 "2025-04-17T03:11:56Z")
It is what is “means”. It’s not at all how it is defined. Open question is how useful it is to the intended audience. There is a middle ground between something that isn’t true at all and going to a delta-epsilon definition. ---
**Author:** ![Francis\_Vaughan](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/francis_vaughan/32/3093_2.png) [@Francis\_Vaughan](https://boards.straightdope.com/u/Francis_Vaughan)\ **Post date:** [April 17, 2025, 2:16pm UTC](https://boards.straightdope.com/t/does-i-to-the-power-of-infinity-0/1017083/25 "2025-04-17T14:16:47Z")
An appeal to a higher authority gets one to Terrence Tao’s missive on the topic. The idea of applying a smoothing function to the series, rather than abrupt termination when reasoning about the series values is IMHO remarkably satisfying. > **[The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and...](https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/#zeta-def)** > > The Riemann zeta function $latex {\\zeta(s)}&fg=000000$ is defined in the region $latex {\\hbox{Re}(s)\>1}&fg=000000$ by the absolutely convergent series $latex \\displaystyle \\zeta(s) = \\su… [Previous page](https://boards.straightdope.com/t/does-i-to-the-power-of-infinity-0/1017083.md?page=1)