Unity is confusing!

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

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.

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, 5 months ago

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

- 3 years, 5 months ago

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

- 3 years, 5 months ago

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

- 3 years, 4 months ago

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-constant function, whereas my $f(x)$ is a constant function of 1.

- 3 years, 4 months ago

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

- 3 years, 4 months ago

- 3 years, 4 months ago

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 $1^\infty=1$ 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 $x$.

- 3 years, 4 months ago

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....

- 3 years, 4 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.

- 3 years, 4 months ago

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?

- 3 years, 5 months ago

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

- 3 years, 5 months 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?

- 3 years, 5 months ago

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

- 3 years, 5 months ago