\documentclass[a4paper]{article}
\usepackage{amsmath}
\begin{document}
\title{Calculate Series $1-\frac14+\frac17-\frac{1}{10}+\cdots$}
\author{Amit Yaron}
\date{Dec.22. 2025}
\maketitle
How do we find the limit
\[
  \sum_{n=0}^{\infty}\frac{(-1)^n}{3n+1}
\]
?\\
First, let us prove that the series converges:
\\
The series is an infinite sum of term with an alternating sign, and the absolute
value of the terms is decreasing and goes to zero as $n$ approaches infinity.
Thus, the series converges.\\
Let us now define the function:
\[
  f(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{3n+1}}{3n+1}
\]
And $f(0)=0$
\[
  f'(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(3n+1)x^{3n}}{3n+1}=\sum_{n=0}^{\infty}(-1)^nx^{3n}
\]
And if we take $y=x^3$, then
\[
  f'(x)=sum_{n=0}^{\infty}(-1)^nx^{3n}=f'(x)=sum_{n=0}^{\infty}(-1)^n{(x^3)}^n=  sum_{n=0}^{\infty}(-1)^ny^n=\frac1{1+y}=\frac1{1+x^3} 
\]
Now, $f'(x)$ will be integrated using partial fractions:
\[
  x^3+1=(x+1)(x^2-x+1)
\]
So, let us find constants $A,B,C$, such that:
\begin{align*}
  \frac1{x^3+1}&=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}\\
               &=\frac{Ax^2-Ax+A+Bx^2+Bx+Cx+C}{x^3+1}\\
               &=\frac{(A+B)x^2+(B-A+C)x+(A+C)}{x^3+1}\\
\end{align*}
Then,
\[
  \begin{cases}
     A+B=0\\
     -A+B+C=0\\
     A+C=1 
  \end{cases}
 \]   
From the first equation:
\[ B=-A \]
Let us plug it into the rest:
\[
  \begin{cases}
    -2A+C=0\\
      A+C=1
  \end{cases}
\]
Substracting the first from the second:
\[
  3A=1\Rightarrow A=\frac13
\]
Thus,
\[ B=-A=-\frac13 \]
and
\[ C=1-A=1-\frac13=\frac23 \]
which means that:
\[
\frac{1}{x^3+1}=\frac13\bigg(\frac1{x+1}\bigg)-\frac13\bigg(\frac{x-2}{x^2-x+1}\bigg)
\]
Let us integrate:
\begin{align*}
  f(x)&=\int f'(x) dx\\
  &=\frac13\int\frac1{x+1}-\frac13\int\frac{x-2}{x^2-x+1}\\
  &=\frac{\ln|x+1|}3-\frac16\int\frac{2x-4}{x^2-x+1}\\
  &=\frac{\ln|x+1|}3-\frac16\int\frac{2x-1-3}{x^2-x+1}\\
  &=\frac{\ln|x+1|}3-\frac16\int\frac{2x-1}{x^2-x+1}dc+\frac16\int\frac3{x^2-x+1}dx\\
  &=\frac{\ln|x+1|}3-\frac{\ln|x^2-x+1|}6+\frac12\int\frac1{x^2-x+1}dx\\
  &=\frac{\ln|x+1|}3-\frac{\ln|x^2-x+1|}6+2\int\frac1{4x^2-4x+4}dx\\
  &=\frac{\ln|x+1|}3-\frac{\ln|x^2-x+1|}6+2\int\frac1{4x^2-4x+1+3}dx\\
  &=\frac{\ln|x+1|}3-\frac{\ln|x^2-x+1|}6+2\int\frac1{(2x-1)^2+3}dx\\
  &=\frac{\ln|x+1|}3-\frac{\ln|x^2-x+1|}6+\frac23\int\frac1{\big(\frac{2x-1}{\sqrt3}\big)^2+1}dx
\end{align*}
Let us substitute
\begin{align*}
  &u=\arctan\big(\frac{2x-1}{\sqrt3}\big)\Rightarrow\\
  \Rightarrow &\frac{2x-1}{\sqrt3}=\tan(u)\Rightarrow\\
  \Rightarrow &2x-1=\sqrt3\tan(u)\Rightarrow\\
  \Rightarrow &x=\frac{1+\sqrt3\tan(u)}2\Rightarrow\\
  \Rightarrow &\frac{dx}{du}=\frac{\sqrt3}2\frac{d}{du}\frac{\sin(u)}{\cos(u)}\Rightarrow\\
  \Rightarrow &\frac{dx}{du}=\frac{\sqrt3}2\bigg(\frac{\cos^2(u)+\sin^2(u)}{\cos^2(u)}\bigg)\Rightarrow\\
  \Rightarrow &\frac{dx}{du}=\frac{\sqrt3}2\bigg(1+\frac{\sin^2(u)}{\cos ^2(u)}\bigg)\Rightarrow\\
  \Rightarrow &\frac{dx}{du}=\frac{\sqrt3}2[1+\tan^2(u)]
\end{align*}
Thus,
\[
  \int\frac1{\big(\frac{2x-1}{\sqrt3}\big)^2+1}dx=\int\frac{\sqrt3}2\bigg(\frac{1+\tan^2(u)}{1+\tan^2(u)}\bigg)=\int\frac{\sqrt3}2du=u+C=\frac{\sqrt3}2\arctan\bigg(\frac{2x-1}{\sqrt3}\bigg)+C
\]
And thus,
\begin{align*}
  f(x)&=\frac{\ln|x+1|}3-\frac{\ln|x^2-x+1|}6+\frac23\cdot\frac{\sqrt3}2\arctan\bigg(\frac{2x-1}{\sqrt3}\bigg)+C\\
  &=\frac{\ln|x+1|}3-\frac{\ln|x^2-x+1|}6+\frac{\sqrt3}3\arctan\bigg(\frac{2x-1}{\sqrt3}\bigg)+C
\end{align*}
Now, it's time to find $C$.\\
Recall that
\begin{align*}
  &f(0)=0\Rightarrow\\
  \Rightarrow &\frac{\ln|0+1|}3-\frac{\ln|0^2-0+1|}6+\frac{\sqrt3}3\arctan\bigg(\frac{2\cdot0-1}{\sqrt3}\bigg)+C=0\Rightarrow\\
  \Rightarrow&\frac{\ln|1|}3-\frac{\ln|1|}6+\frac{\sqrt3}3\arctan\bigg(\frac{-1}{\sqrt3}\bigg)+C=0\Rightarrow\\
  \Rightarrow&\frac{\sqrt3}3\arctan\bigg(\frac{-1}{\sqrt3}\bigg)+C=0\Rightarrow\\
  \Rightarrow&C=\frac{\sqrt3}3\arctan\bigg(\frac{1}{\sqrt3}\bigg)\Rightarrow\\
  \Rightarrow&C=\frac{\sqrt3}3\arctan\bigg(\frac{\frac12}{\frac{\sqrt3}2}\bigg)=\frac{\sqrt3}3\arctan\bigg(\frac{\sin\big(\frac{\pi}6\big)}{\cos\big(\frac{\pi}6\big)}\bigg)=\frac{\sqrt3}3\arctan\bigg(\tan\big(\frac{\pi}6\big)\bigg)\Rightarrow\\
  \Rightarrow&C=\frac{\sqrt3\pi}{18}\Rightarrow\\
  \Rightarrow&f(x)=\frac{\ln|x+1|}3-\frac{\ln|x^2-x+1|}6+\frac{\sqrt3}3\arctan\bigg(\frac{2x-1}{\sqrt3}\bigg)+\frac{\sqrt3\pi}{18}
\end{align*}
Finally,
\begin{align*}
  \sum_{n=0}^{\infty}\frac{(-1)^n}{3n+1}&=f(1)\\
  &= \frac{\ln|1+1|}3-\frac{\ln|1^2-1+1|}6+\frac{\sqrt3}3\arctan\bigg(\frac{2\cdot1-1}{\sqrt3}\bigg)+\frac{\sqrt3\pi}{18}\\
  &= \frac{\ln(2)}3-\frac{\ln(1)}6+\frac{\sqrt3}3\cdot\frac{\pi}6+\frac{\sqrt3\pi}{18}\\
  &= \frac{\ln(2)}3+\frac{\sqrt3\pi}{18}+\frac{\sqrt3\pi}{18}\\
  &= \frac{\ln(2)}3+\frac{\sqrt3\pi}9
\end{align*}
\end{document}