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

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 \(Not\) \(defined\).

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.

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## Comments

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

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

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

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

\(\therefore\) As \(x\to\infty\), \(\dfrac{1}{x}\to 0\), hence limit of the function reaches \(1\).

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

Not defined.But if we follow \(\zeta\left(-1\right)=\dfrac{-1}{12}\), then this turns out to be \(\large1^{\frac{-1}{12}}\) which I'm confused about because the \(12\)th root of \(1\) is both \(1\) and \(-1\) (or is it just 1?) and \(10\) complex roots.

So, what can be the answer?

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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

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

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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.

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

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It's correct. The link you gave me have the limit in the form of \((f(x))^{g(x)} \), where \(f(x) \) is a

non-constantfunction, whereas my \(f(x) \)isa constant function of 1.Log in to reply

here's a different link to support my statement. Also, this.

Well then,Log in to reply

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

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

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

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