# General Method for Summing Divergent Series. Determination of Limits of Divergent Sequences and Functions in Singular Points v2

Abstract. In this work I am going to mention historical development of divergent series theory, and to give a number of different examples, as some of the methods for their summing. After that, I am going to introduce the general method, which I discovered, for summing divergent series, which we can also consider as a method for computing limits of divergent sequences and functions in divergent points, In this case, limits of sequences of their partials sums. Through the exercises, I am going to apply this method on given examples and prove its validity. Then I’m going to apply the method to compute the value of some divergent integrals.

— Algorithmization of the method —

Function ${f,g}$; EquationSolver solveEquation;

$\displaystyle n\in N;$

$\displaystyle \lim^D_{z\rightarrow+\infty} z^n=\int_{-1}^0 z^n dz;$

$\displaystyle \lim^D_{z\rightarrow-\infty} z^n=\int_0^1 z^n dz;$

$\displaystyle \lim^D_{z\rightarrow 0^+} z^{-n}=\int_{-\infty}^{-1} z^{-n-2} dz;$

$\displaystyle \lim^D_{z\rightarrow 0^-} z^{-n}=\int_1^{+\infty} z^{-n-2} dz;$

$\displaystyle a, const\in C\bigcup\{\infty\},\ \alpha\in[0,2\pi);$

if ((${f}$ has a pole at ${a}$)||(${f}$ has a removable singularity at ${a}$)) {

if ((${g}$ has a pole at ${a}$)||(${g}$ has a removable singularity at ${a}$)) {

$\displaystyle \lim^D_{z\rightarrow a(\alpha)^\pm}(f(z)+g(z))=\lim^D_{z\rightarrow a(\alpha)^\pm}f(z)+\lim^D_{z\rightarrow a(\alpha)^\pm}g(z);$

$\displaystyle \lim^D_{z\rightarrow a(\alpha)^\pm}(c\cdot f(z))=c\cdot\lim^D_{z\rightarrow a(\alpha)^\pm}f(z);$

}

if (${f}$ has a removable singularity at ${a}$) {

$\displaystyle \lim^D_{z\rightarrow a} f(z)=\lim^D_{z\rightarrow a(\alpha)^\pm} f(z)=\lim_{z\rightarrow a} f(z);$

}

if (${f}$ has a pole at ${a}$) {

if (${a==\infty}$) {

$\displaystyle f(z)=F_1(z)+F_2(z)=\sum_{k=-n}^{-1} c_k\cdot\frac{1}{z^k}+\sum_{k=0}^\infty c_k\cdot\frac{1}{z^k};$

$\displaystyle \lim^D_{z\rightarrow\infty(\alpha)^+}f(z)=\lim^D_{z\rightarrow+e^{i\alpha}\infty}f(z)=\lim^D_{r\rightarrow+\infty}f(r\cdot e^{i\alpha});$

$\displaystyle \lim^D_{z\rightarrow\infty(\alpha)^-}f(z)=\lim^D_{z\rightarrow-e^{i\alpha}\infty}f(z)=\lim^D_{r\rightarrow-\infty}f(r\cdot e^{i\alpha});$

$\displaystyle \lim^D_{z\rightarrow\infty(\alpha)}f(z)=\lim^D_{r\rightarrow\infty(0)}f(r\cdot e^{i\alpha});$

$\displaystyle \lim^D_{z\rightarrow\infty(\alpha)}f(z)=\frac{1}{2}(\lim^D_{z\rightarrow\infty(\alpha)^+}f(z)+\lim^D_{z\rightarrow\infty(\alpha)^-}f(z));$

$\displaystyle \lim^D_{z\rightarrow\infty}f(z)=\frac{1}{2\pi}\int_0^{2\pi}\lim^D_{z\rightarrow\infty(\alpha)}f(z)d\alpha;$

$\displaystyle \lim^D_{x\rightarrow\infty(0)^+}f(x)=\lim^D_{x\rightarrow+\infty}f(x);$

$\displaystyle \lim^D_{x\rightarrow\infty(0)^-}f(x)=\lim^D_{x\rightarrow-\infty}f(x);$

$\displaystyle \lim^D_{z\rightarrow \infty(\alpha)\pm}f(z)=\sum_{k=1}^{n} c_{k}\cdot e^{i\alpha k}\frac{(\pm 1)^k}{2(k+1)}+c_0;$

$\displaystyle \lim^D_{z\rightarrow \infty(\alpha)}f(z)=\sum_{k=1}^n c_k\cdot e^{i\alpha k}\frac{1+(-1)^k}{2(k+1)}+c_0;$

}

if (${a\in C}$) {

$\displaystyle f(z)=F_1(z)+F_2(z)=\sum_{k=-n}^{-1} c_k\cdot(z-a)^k+\sum_{k=0}^\infty c_k\cdot(z-a)^k;$

$\displaystyle \lim^D_{z\rightarrow a(\alpha)^+}f(z)=\lim^D_{z\rightarrow a+e^{i\alpha}0}f(z)=\lim^D_{r\rightarrow 0^+}f(a+r\cdot e^{i\alpha});$

$\displaystyle \lim^D_{z\rightarrow a(\alpha)^-}f(z)=\lim^D_{z\rightarrow a-e^{i\alpha}0}f(z)=\lim^D_{r\rightarrow 0^-}f(a+r\cdot e^{i\alpha});$

$\displaystyle \lim^D_{z\rightarrow a(\alpha)}f(z)=\lim^D_{r\rightarrow 0(0)}f(a+r\cdot e^{i\alpha});$

$\displaystyle \lim^D_{z\rightarrow a(\alpha)}f(z)=\frac{1}{2}(\lim^D_{z\rightarrow a(\alpha)^+}f(z)+\lim^D_{z\rightarrow a(\alpha)^-}f(z));$

$\displaystyle \lim^D_{z\rightarrow a}f(z)=\frac{1}{2\pi}\int_0^{2\pi}\lim^D_{z\rightarrow a(\alpha)}f(z)d\alpha;$

$\displaystyle \lim^D_{x\rightarrow a(0)^+}f(x)=\lim^D_{x\rightarrow a^+}f(x);$

$\displaystyle \lim^D_{x\rightarrow a(0)^-}f(x)=\lim^D_{x\rightarrow a^-}f(x);$

$\displaystyle \lim^D_{z\rightarrow a(\alpha)^\pm}f(z)=\sum_{k=-n}^{-1} c_{k}\cdot e^{i\alpha k}\frac{(\pm 1)^k}{2(-k+1)}+c_0;$

$\displaystyle \lim^D_{z\rightarrow a(\alpha)}f(z)=\sum_{k=-n}^{-1} c_{k}\cdot e^{i\alpha k}\frac{1+(-1)^k}{2(-k+1)}+c_0;$

}

$\displaystyle \lim^D_{z\rightarrow a}f(z)=c_0;$

}

} else if((${\lim^D_{z\rightarrow a(\alpha)\pm}f(z)}$ is not finite)||(${\lim^D_{z\rightarrow a(\alpha)\pm}f(z)}$ does not exist)) { if((${f}$ can be expanded into a generalized Puiseux series about point ${a}$)&&

(!(${f(z)=\sum_{k=-n}^{+\infty}c_k(z)\cdot \frac{1}{z^k}}$)!=!(${f(z)=\sum_{k=-n}^{+\infty}c_k(z)\cdot (z-a)^k)}$)) {

${const}$=solveEquation.forConstantOfIntegration(${\int d(c_0(z))=c_0(z)}$);

$\displaystyle \lim^D_{z\rightarrow a(\alpha)\pm}f(z)=const;$

} else {

$\displaystyle c_0(z)=f(z);$

${const}$=solveEquation.forConstantOfIntegration(${\int d(c_0(z))=c_0(z)}$);

$\displaystyle \lim^D_{z\rightarrow a(\alpha)\pm}f(z)=const;$

}

$\displaystyle \lim^D_{z\rightarrow a}f(z)=const;$

}

— The sum of some divergent series —

— Determination of values of some divergent integrals —