6
votes
0answers
154 views

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 ...
1
vote
1answer
105 views

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. $$1+1+1+1+1+\ldots=\zeta(0)=-\frac{1}{2}$$ $$1+2+3+4+5+\ldots=\zeta(-1)=-\frac{1}{12}$$ ...
4
votes
2answers
199 views

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 ...
-1
votes
1answer
37 views

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 ...
5
votes
4answers
188 views

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: ...
6
votes
2answers
288 views

Problem in Divergent series

Lets try to evaluate $$\frac{(-1)}{1^s}+\frac{(-1)^2}{2^s}+\frac{(-1)^3}{3^s}+...$$ $$=\frac{e^{\pi i}}{1^s}+\frac{e^{2\pi i}}{2^s}+\frac{e^{3\pi i}}{3^s}+...$$ $$=\frac{1}{1^s}(1+\frac{\pi ...
6
votes
3answers
723 views

$1 + 1 + 1 +\cdots = -\frac{1}{2}$

The formal series $$ \sum_{n=1}^\infty 1 = 1+1+1+\dots=-\frac{1}{2} $$ comes from the analytical continuation of the Riemann zeta function $\zeta (s)$ at $s=0$ and it is used in String Theory. I am ...
1
vote
1answer
56 views

Series divergence - how precise should the answer be

Morning. I've written down some of my reasoning and arguments as to why the series diverges, however I am not certain I can safely conclude it diverges to $\infty$. Would you give it a look, please? ...
10
votes
0answers
359 views

I can Euler-sum $\sqrt{-\ln(1)}-\sqrt{-\ln(2)}+\sqrt{-\ln(3)}-\cdots$. But how can I do $\sqrt{-\ln(1)}+\sqrt{-\ln(2)}+\sqrt{-\ln(3))}+\cdots$?

This is also related to an older thread in MSE ("what is the half derivative of zeta at zero?") . One of the possible steps in the problem of that thread was to evaluate the series ...
1
vote
0answers
429 views

Ramanujan Summation not consistent with Riemann's Zeta function?

Wikipedia states that Ramanujan sums and the Riemann Zeta function have the same values for even $k$: $$1 + 2^{2k} + 3^{2k} + \cdots = 0\ (\Re)$$ However, I don't understand how this can be true, ...
2
votes
1answer
114 views

What's the lowest real $x$ such that $\zeta(x)$ converges?

It's easy to prove that$$\zeta(1)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...$$ diverges, and $$\zeta(2)=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...$$ converges to $\frac{\pi^2}{6}$. Intuiting the ...
5
votes
0answers
178 views

On the series $ \displaystyle \sum_{n=1}^{\infty} \frac{\Lambda(n)}{\sqrt{n}} \cos(x \log(n) + a) $.

Is the following series ‘summable’ in the sense that it may be divergent but we can attach a meaning to it? $$ \sum_{n=1}^{\infty} \frac{\Lambda(n)}{\sqrt{n}} \cos(x \log(n) + a) = (\text{reg}) ...
2
votes
1answer
165 views

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 ...
2
votes
2answers
221 views

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 ...
88
votes
8answers
15k views

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?
9
votes
2answers
809 views

Values of the Riemann Zeta function and the Ramanujan Summation - How strong is the connection?

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 ...