# Tagged Questions

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### Connection between integral expression and the factorial of infinity

Does the fact that $$\int_{-\infty}^{\infty}\exp\left(-\frac{1}{2}x^2\right)\mathrm{d}x=\sqrt{2\pi}$$ Have something to do with the fact that the regularized factorial of infinity is also ...
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### Playing fast and loose with divergent series [closed]

I have been playing around recently with the regularization of infinite divergent sums and products, e.g. $$1+1+1+1+1+\ldots=\zeta(0)=-\frac{1}{2}$$ $$1+2+3+4+5+\ldots=\zeta(-1)=-\frac{1}{12}$$ ...
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### Sum of divergent series

I saw a lot of article in Math SE like Why does 1+2+3+⋯=−1/12? and S=1+10+100+100+10000+…=−1/9? How is that and lot of others. Also I saw this one of Ramanujan summation but I do not get the ...
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### Is there an expression for $(S/k)$ where $S=\sum_{n=1}^\infty n$ and $k \in \mathbb{Z}$?

Given that $S=\sum_{n=1}^\infty n=-1/12$ (for an explanation see this question or this video from Youtube) For example if $k=4$: $(S/4)=1/4+2/4+3/4+1+5/4+6/4+7/4+2+9/4...$ Please edit to improve ...
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### Find the regularized sum $\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+…$

By considering the integral Zeta function $$F(s)=s+\frac{1}{2^s\ln(2)}+\frac{1}{3^s\ln(3)}+\frac{1}{4^s\ln(4)}+...$$ Evaluate $$\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+...$$ EDIT: ...
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### Meaning of equality in zeta regularization

It is known that $$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = \infty$$ but it is also known that $$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = -\frac{1}{{12}}$$ which can obtained ...
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### Sum of the Stieltjes constants? (divergent summation)

The sequence of Stieltjes-constants diverges and thus cannot be summed conventionally. However their signs oscillate (unfortunately non-periodic) and thus I tried Euler- and a version of ...
### Why does $1+2+3+\dots = -\frac{1}{12}$?
$\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}$ only converges to $\zeta(s)$ if $\text{Re}(s) > 1$. Why should analytically continuing to $\zeta(-1)$ give the right answer?
The Ramanujan Summation of some infinite sums is consistent with a couple of sets of values of the Riemann zeta function. We have, for instance, \zeta(-2n)=\sum_{n=1}^{\infty} n^{2k} = 0 ...