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

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.

## Comments

Sort by:

TopNewestWe are essentially calculating \( \displaystyle \lim_{n\to\infty} 1^{n(n+1)/2} = 1^\infty = 1 \). – Pi Han Goh · 2 months, 3 weeks ago

Log in to reply

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

\(\because\) 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? – Tapas Mazumdar · 2 months, 3 weeks ago

Log in to reply

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

Read upPlus, 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 · 2 months, 3 weeks ago

Log in to reply

\(1^\infty\) is indeterminate. – Prasun Biswas · 2 months ago

Even though the limit equals 1, the equality \(1^\infty=1\) you mention is incorrect sinceLog in to reply

non-constantfunction, whereas my \(f(x) \)isa constant function of 1. – Pi Han Goh · 2 months agoLog in to reply

here's a different link to support my statement. Also, this. – Prasun Biswas · 2 months ago

Well then,Log in to reply

Log in to reply

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 \(x\). – Prasun Biswas · 2 months ago

Log in to reply

Is it just me, but I tried editing my comment but I can't seem to do it. Weird.... – Pi Han Goh · 2 months ago

Log in to reply

– Prasun Biswas · 2 months ago

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.Log in to reply

– Tapas Mazumdar · 2 months, 3 weeks ago

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?Log in to reply

sums of divergent series. – Pi Han Goh · 2 months, 3 weeks ago

Think about how zeta(-1) = -1/12 was derived in the first place. Did it apply Abel sum? Read upLog in to reply

all i see is an infinite number of 1's being multiplied. – Frank Giordano · 2 months, 2 weeks ago

Log in to reply