# Is there a summation method where the divergent series $S = U_0+U_1+U_2+\dots$ converges to a finite value(V2)?

I have a question regarding this question here.

is-there-a-summation-method-where-the-divergent-series

if I set $$x+2=c/c-v$$ , will I have
$$U_n = M\left(c-\frac{c}{n+2}\right)-M\left(c-\frac{c}{n+1}\right), \label{1}\tag{U}$$

and will $$M(c)=M(c-1)$$, or $$M(c)=-M(c-c)=-M(0)=-m_0$$?

• What question? Your link just tries to go to the general list mathoverflow.net/questions. Commented Jul 26, 2023 at 1:49
• Sorry, it's okay now Commented Jul 26, 2023 at 1:58

Are the $$U_n$$ supposed to have a physical meaning or is this a purely mathematical exercises? if we directly interpret:
$$U_n = M\left(c-\frac{1}{n+2}\right)-M\left(c-\frac{1}{n+1}\right)\label{2}\tag{U} = \frac{m_0}{\sqrt{1 - 1 + \frac{2}{c(n+2)} - \frac{1}{c^2(n+2)^2}}} - \frac{m_0}{\sqrt{1 - 1 + \frac{2}{c(n+1)} - \frac{1}{c^2(n+1)^2}}} = m_0 \left( \frac{1}{\sqrt{\frac{2}{c(n+2)} - \frac{1}{c^2(n+2)^2}}} - \frac{1}{\sqrt{\frac{2}{c(n+1)} - \frac{1}{c^2(n+1)^2}}} \right) = m_0 c \left( \frac{n+2}{\sqrt{2c(n+2) -1 }} - \frac{n+1}{\sqrt{2c(n+1)-1}} \right) =$$
It's not clear to me how to assign units to the last object. It seems like the $$n$$ and $$1$$ and $$2$$ need to be given units for the whole thing to remain physical with dimensional analysis.
Anyways... I believe that $$U_n = O(n^{-\frac{1}{2}})$$ so one could use the Euler Maclaurin formula on this and see if that suggests a constant term in its asymptotic expansion. This will be very tedious and I'll get around to it later (unless someone beats me to it :) )
• Thank you very much. It would be a physically sound theory formulated mathematically in that case. Have you considered that $x+2=\frac{1}{1-\frac{v^2}{c^2}}$? Commented Jul 25, 2023 at 20:38