Used to approximate function \(f\relax (x) \) at some \(x\) knowing its values and all its derivatives at some point \(x_{0}\), called the expansion point.\[ f\relax (x) =f\left (x_{0}\right ) +\left (x-x_{0}\right ) f^{\prime }\left (x_{0}\right ) +\frac {1}{2}\left (x-x_{0}\right ) ^{2}f^{\prime \prime }\left (x_{0}\right ) +\cdots \]\begin {align*} \sin x & =x-\frac {x^{3}}{3!}+\frac {x^{5}}{5!}+\cdots \\ \cos x & =1-\frac {x^{2}}{2!}+\frac {x^{4}}{4!}-\cdots \end {align*}
To find series for \(\ln \left (1+x\right ) \), do this \begin {align*} \int \frac {1}{1+x}dx & =\ln \left (1+x\right ) +C\\ \int (1-x+x^{2}-x^{3}+\cdots )dx & =\ln \left (1+x\right ) +C\\ x-\frac {x^{2}}{2}+\frac {x^{3}}{3!}-\cdots & =\ln \left (1+x\right ) +C\qquad \left \vert x\right \vert <1 \end {align*}
To find \(C\), let \(x=0\). Hence \(0=\ln \relax (1) +C\). So \(C=-\ln \left ( 1\right ) \). Therefore\[ \ln \left (1+x\right ) =\ln \relax (1) +x-\frac {x^{2}}{2}+\frac {x^{3}}{3!}-\cdots \qquad \left \vert x\right \vert <1 \] And\begin {align*} \int \frac {1}{1-x}dx & =-\ln \left (1-x\right ) +C\\ -\int (1+x+x^{2}+x^{3}+\cdots )dx & =\ln \left (1-x\right ) +C\\ -\left (x+\frac {x^{2}}{2}+\frac {x^{3}}{3}+\cdots \right ) & =\ln \left ( 1-x\right ) +C\\ -x-\frac {x^{2}}{2}-\frac {x^{3}}{3}+\cdots & =\ln \left (1-x\right ) +C\qquad \left \vert x\right \vert <1 \end {align*}
To find \(C\), let \(x=0\). Hence \(0=\ln \relax (1) +C\). So \(C=-\ln \left ( 1\right ) \). Therefore\[ \ln \left (1-x\right ) =\ln \relax (1) -x-\frac {x^{2}}{2}-\frac {x^{3}}{3!}+\cdots \] And \(\ln \left (1+2x\right ) \) series is found as follows\begin {align*} \int \frac {1}{1+2x}dx & =\frac {1}{2}\ln \left (1+2x\right ) +C\\ \int (1-2x+\left (2x\right ) ^{2}-\left (2x\right ) ^{3}+\cdots )dx & =\frac {1}{2}\ln \left (1+2x\right ) +C\\ \left (x-\frac {2x^{2}}{2}+\frac {4x^{3}}{3}-\frac {8x^{4}}{4}\cdots \right ) & =\frac {1}{2}\ln \left (1+2x\right ) +C\qquad \left \vert x\right \vert <1 \end {align*}
To find \(C\), let \(x=0\). Hence \(0=\ln \relax (1) +C\). So \(C=-\ln \left ( 1\right ) \). Therefore\[ \ln \left (1+2x\right ) =2\ln \relax (1) +2\left (x-\frac {2x^{2}}{2}+\frac {4x^{3}}{3}-\frac {8x^{4}}{4}\cdots \right ) \] And\begin {align*} e^{x} & =1+x+\frac {x^{2}}{2!}+\frac {x^{3}}{3!}+\cdots =\sum _{n=0}^{\infty }\frac {x^{n}}{n!}\\ \tan x & =x+\frac {x^{3}}{3}+\frac {2}{15}x^{5}+\cdots \end {align*}
Some others \begin {align*} \frac {1}{1+x} & =1-x+x^{2}-x^{3}+\cdots \qquad \left \vert x\right \vert <1\\ \frac {1}{1-x} & =1+x+x^{2}+x^{3}+\cdots \qquad \left \vert x\right \vert <1\\ \left (1+x\right ) ^{a} & =\sum \binom {a}{n}x^{n} \end {align*}
Where \(\binom {a}{n}\) is binomial coefficient \(\binom {a}{n}=\frac {a!}{n!\left ( a-n\right ) !}\). General Binomial\[ \left (1+x\right ) ^{n}=1+nx+\frac {n\left (n-1\right ) }{2!}x^{2}+\frac {n\left (n-1\right ) \left (n-2\right ) }{3!}x^{3}+\cdots \] This works for positive and negative \(n\), rational or not. The sum converges only for \(\left \vert x\right \vert <1\). So, for \(n=-1\) the above becomes\[ \frac {1}{\left (1+x\right ) }=1-x+x^{2}-x^{3}+\cdots \] And\[ \frac {1}{\left (1-x\right ) ^{2}}=\sum _{n=1}^{\infty }nx^{n-1}=1+2x+3x^{2}+4x^{3}+\cdots \] And\[ \left (1+x\right ) ^{p}=1+px+p\left (p-1\right ) x^{2}\cdots \] For small \(x\) the above approximates to\[ \left (1+x\right ) ^{p}=1+px \]
First test, check if \(\lim _{n\rightarrow \infty }a_{n}\) goes to zero. If not, then no need to do anything. Series does not converge. Then use ratio test. If\[ \lim _{n\rightarrow \infty }\left \vert \frac {a_{n+1}}{a_{n}}\right \vert <1 \] Then converges. if result is \(>1\) then diverges. If result is one, then more testing is needed. If converges, then radius of convergence \(R\) is\begin {align*} R & =\lim _{n\rightarrow \infty }\left \vert \frac {a_{n}}{a_{n+1}}\right \vert \\ \left \vert x\right \vert & <R \end {align*}
\begin {align*} \sum _{n=1}^{N}n & =\frac {1}{2}N\left (N+1\right ) \\ \sum _{n=1}^{N}a_{n} & =N\left (\frac {a_{1}+a_{N}}{2}\right ) \end {align*}
i.e. the sum is \(N\) times the arithmetic mean.
Geometric series. \begin {align*} S & =a+ar+ar^{2}+ar^{3}+\cdots \\ & =\sum _{k=0}^{N}ar^{k}\\ & =a\left (\frac {1-r^{N+1}}{1-r}\right ) \end {align*}
For \(\left \vert r\right \vert <1\)\[ S=\frac {a}{1-r}\]