Unity is confusing!

11+2+3+4+5+6+7\huge 1^{1+2+3+4+5+6+7\cdots}

Considering the fact that some diverging sums can also approach a certain limit. What is the sum of real value(s) of the expression above?

Details:

  • If no real value(s) are obtained give your answer as NotNot defineddefined.

  • If the limit approaches \infty then enter your answer as \infty as well.

About:

This problem is original. Upon pondering over the answer to this problem I wasn't able to come up with a legit explanation as to why my method was/wasn't correct. Please enter your answer with an appropriate explanation.

Note by Tapas Mazumdar
3 years, 1 month ago

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We are essentially calculating limn1n(n+1)/2=1=1 \displaystyle \lim_{n\to\infty} 1^{n(n+1)/2} = 1^\infty = 1 .

Pi Han Goh - 3 years, 1 month ago

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Also for any real number nn (say) we can say that:

limxnx=1\displaystyle \lim_{x\to\infty}{\large\sqrt[x]{n}} = 1

\because The above can be written as limxn1x\displaystyle \lim_{x\to\infty}{\large n^{\frac{1}{x}}}.

\therefore As xx\to\infty, 1x0\dfrac{1}{x}\to 0, hence limit of the function reaches 11.

So, if n=1\sqrt[\infty]{n}=1 (as in most cases, the limit and the value mean quite the same thing), can't we say that 1=n1^{\infty} = n, which is kind of a paradox. And the only plausible explanation to this is that 11^{\infty} is Not defined.

But if we follow ζ(1)=112\zeta\left(-1\right)=\dfrac{-1}{12}, then this turns out to be 1112\large1^{\frac{-1}{12}} which I'm confused about because the 1212th root of 11 is both 11 and 1-1 (or is it just 1?) and 1010 complex roots.

So, what can be the answer?

Tapas Mazumdar - 3 years, 1 month ago

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Read up indetermediate forms, you did not obey those rules, so your logic is incorrect.

Plus, if you want to invoke 1+2+ 3 + ... = -1/12, then you should make it clear that you're using riemann zeta regularization from the start. Otherwise, by convention, 1 +2 + 3 + ... = infinity

Pi Han Goh - 3 years, 1 month ago

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But the string theory tells us that n=1n (or) ζ(1)=112\displaystyle \sum_{n=1}^{\infty}{n}~ \left(\text{or}\right)~ \zeta\left(-1\right) = \dfrac{-1}{12}. Wouldn't that concept be counted right here?

Tapas Mazumdar - 3 years, 1 month ago

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Think about how zeta(-1) = -1/12 was derived in the first place. Did it apply Abel sum? Read up sums of divergent series.

Pi Han Goh - 3 years, 1 month ago

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Even though the limit equals 1, the equality 1=11^\infty=1 you mention is incorrect since 11^\infty is indeterminate.

Prasun Biswas - 3 years ago

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It's correct. The link you gave me have the limit in the form of (f(x))g(x)(f(x))^{g(x)} , where f(x)f(x) is a non-constant function, whereas my f(x)f(x) is a constant function of 1.

Pi Han Goh - 3 years ago

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@Pi Han Goh Well then, here's a different link to support my statement. Also, this.

Prasun Biswas - 3 years ago

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@Prasun Biswas

Pi Han Goh - 3 years ago

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@Pi Han Goh I did. I never said that the limit doesn't equals 1. In fact, my first reply to you on this thread states "Even though the limit equals 1, the equality..."

I'm disputing the parts 1=11^\infty=1 and limn1n(n+1)/2=1\lim\limits_{n\to\infty}1^{n(n+1)/2}=1^\infty in your comment. You should write it more like limn1n(n+1)/2=limn1=1\lim\limits_{n\to\infty}1^{n(n+1)/2}=\lim\limits_{n\to\infty}1=1 where the second equality follows from the fact that 1x=11^x=1 for any real xx.

Prasun Biswas - 3 years ago

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@Prasun Biswas Ah sorry, I didnt' see what I wrote. It's been a long time. Good eye!

Is it just me, but I tried editing my comment but I can't seem to do it. Weird....

Pi Han Goh - 3 years ago

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@Pi Han Goh Yes, that's a known bug. Fortunately, there's a way around it if you're on PC. I posted the method in #bug-report of Brilliant Lounge a few days ago. You can check it out if you want.

Prasun Biswas - 3 years ago

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all i see is an infinite number of 1's being multiplied.

Frank Giordano - 3 years ago

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