Is there a summation method for the divergent series arising from relative mass?

Consider the function $ M(v) = \frac{m_0}{\sqrt{1 - \left(\frac{v}{c}\right)^2}},$ where $v \in \left]-c;c\right[$, $m_0\in\mathbb{R}^{*+}$, and $c=3.10^8$.

* Let $(U_n)$ be a sequence with general term   $U_n = M\left(c-\frac{1}{n+2}\right)-M\left(c-\frac{1}{n+1}\right) $ where $ n \in \mathbb{N}$

* let the sequence its partial sums $S_n$ be defined as $S_n = U_0+U_1+U_2+\ldots+U_n$, and finally

* let $S$ be the sum of the series defined by $U$, i.e. $S = U_0+U_1+U_2+\ldots$

The series $S = U_0+U_1+U_2+\ldots=+\infty$ diverges to infinity and is not Cesàro summable, i.e. it does not converge in the sense of Cesàro, just like the series $1+2+3+\ldots=+\infty$

for more details, please refer to https://en.wikipedia.org/wiki/Cesàro_summation

On the other hand, the divergent series $1+2+3+\dots$ can be summed by using the zeta regularization summation method to $-1/12$

for more details, please   refer to https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF

For your information, $S$ represents the divergence of $M(v)$ at $v=c $

because at $v=c$, we have $M(c)=\lim_{v \to c-} M(v)$.

I let $x+2=1/(c-v)$, so $v=c-1/(x+2)$ Consequently:

$M(c)=\lim_{v \to c-} M(v)=\lim_{{x \to +\infty}}M(c-1/(x+2))$.

Thus, we have:

$M(c)=\lim_{v \to c-} M(v)=\lim_{{x \to +\infty}}M(c-1/(x+2))=\lim_{{n\to +\infty}}M(c-1/(n+2))$.

I let  $S_n=M(c-1/(n+2))$ so  $U_n=S_n-S_{n-1}$ so $S_n = U_0+U_1+U_2+\ldots+U_n$

and $S = U_0+U_1+U_2+\ldots$.

Hence, we have:

$M(c)=\lim_{v \to c-} M(v)=\lim_{{n\to +\infty}}S_n=S$.

Therefore $S$ indeed represents the divergence of $M(v)$ at $v=c $.


Questions:

* Is there a summation method where the divergent series $S = U_0+U_1+U_2+\ldots$ converges to a finite value?

* Is there a summation method where the divergent series $S = U_0+U_1+U_2+\ldots$ converges to a finite value equal to $m_0$? 

Edit:
Here is the solution to the first question:



you can just separate off the divergent part of $M$ and define it via $\zeta $-function regularization. Define $$ M_d (n) = m_0 \sqrt{\frac{c}{2}} \sqrt{n} $$ and write $$ U_n = \left[ M\left( c-\frac{1}{n+2} \right) - M_d (n+2) - M\left( c-\frac{1}{n+1} \right) + M_d (n+1)\right] \\ + [M_d (n+2)] - [M_d (n+1)] $$ and sum the 3 square brackets separately. Note that the combination of the first two terms in the first square brackets vanishes for $n\rightarrow \infty $, and therefore, since the sum over the first square brackets telescopes, that sum reduces to $$ S^{(1)} = -M(c-1) + M_d (1) $$ The sum over the second square brackets is $$ S^{(2)} = m_0 \sqrt{\frac{c}{2} } \sum_{n=2}^{\infty } \sqrt{n} = m_0 \sqrt{\frac{c}{2} } \left( \zeta \left(-\frac{1}{2} \right) - 1 \right) $$ and analogously for the third square brackets $$ S^{(3)} = m_0 \sqrt{\frac{c}{2} } \ \zeta \left(-\frac{1}{2} \right) $$
So the entire sum reduces to the finite value :

$$ \begin{align} S & = S^{(1)} + S^{(2)} - S^{(3)} \\[8pt] & = -M(c-1) + M_d (1) - m_0 \sqrt{\frac{c}{2} } \\[8pt] & =-M(c-1) \end{align}$$

The divergence of $M(v)$ at $v=c $ :     $M(c)=-M(c-1)$


Could you then answer question 2 :

Is there a summation method where the divergent series $S = U_0+U_1+U_2+\ldots$ converges to a finite value equal to $m_0$?
Sequences and Series
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2 Answers

The series you've provided, represented as:

Σ (from n=1 to ∞) [M(c - (n+2)^(-1)) - M(c - (n+1)^(-1))]

where M(v) = 1 - (c/v)^2, c = 3.10^8, and n ∈ ℕ, represents a divergent series. As you have demonstrated, this series does not converge.

To address your questions:

Is there a summation method where the divergent series Σ [M(c - (n+2)^(-1)) - M(c - (n+1)^(-1))] converges to a finite value?
In general, a divergent series cannot be summed to a finite value using standard summation methods. The methods you have referred to in your explanation, such as Cesàro summation or zeta regularization, can sometimes assign finite values to certain divergent series but are not applicable to all divergent series. The convergence of a series depends on its properties, and not all divergent series have a meaningful finite value associated with them.

Is there a summation method where the divergent series Σ [M(c - (n+2)^(-1)) - M(c - (n+1)^(-1))] converges to a finite value equal to �0m0?
As mentioned earlier, standard summation methods cannot be used to obtain finite values for divergent series in general. While some regularization methods like zeta regularization can assign finite values to certain divergent series, they may not yield specific values such as �0m0 for the given series.

The properties of a series and the chosen summation method play crucial roles in determining whether a divergent series can be assigned a finite value or not. For the specific series you've provided, further analysis and research would be required to determine if it can be associated with a specific finite value using more advanced summation techniques or regularization methods.

Archie Mccarthy Archie Mccarthy
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 Holy Cow! Such a huge question for free? who will solve it for free?

Akam Singh Akam Singh
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