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A sequence is simply a list of numbers written in a specific order and is often noted in the format ${\left\{a_{{n}}\right\}_{{n=1}}^{{\infty }}}$. The partial sums, ${s_{{n}}}$, of a sequence are:
\begin{equation*}
{s_{{1}}=a_{{1}}}
\end{equation*}
\begin{equation*}
{s_{{2}}=a_{{1}}+a_{{2}}}
\end{equation*}
\begin{equation*}
{s_{{3}}=a_{{1}}+a_{{2}}+a_{{3}}}
\end{equation*}
\begin{equation*}
{s_{{n}}=a_{{1}}+a_{{2}}+a_{{3}}+\text{.}\text{.}\text{.}+a_{{n}}}
\end{equation*}
We can treat ${a_{{n}}}$ as a function and find its limit to determine whether the sequence converges or diverges, the following rules apply:
The sequence is convergent if ${\underset{{n\rightarrow \infty }}{{\text{lim}}}a_{{n}}}$ exists and is finite
The sequence is divergent if ${\underset{{n\rightarrow \infty }}{{\text{lim}}}a_{{n}}}$ does not exist or is infinite
We can refer to the partial sums of a sequence in terms of an infinite series denoted as:
\begin{equation*}
{\overset{{n}}{\underset{{i=1}}{\sum }}{a_{{i}}}}
\end{equation*}
It is important to note that a sequence is a list of numbers, whereas, a series is a single number.
An infinite series can be said to be convergent if the sequence of partial sums is convergent and its limit is finite. On the contrary, if no limit exists for the sequence, then the infinite series is divergent.
As a whole, finding an equation for the general term of a series is extremely complicated, so we will pass up all the proofs and cover only the most common series and their terms:
General:
Sum of the first n numbers:
\begin{equation*}
{\sum {n}=1+2+3+4\text{.}\text{.}\text{.}+n=\frac{n(n+1)}{2}}
\end{equation*}
Sum of the squares of the first n numbers:
\begin{equation*}
{\sum {n}^{{2}}=1^{{2}}+2^{{2}}+3^{{2}}+4^{{2}}\text{.}\text{.}\text{.}+n^{{2}}=\frac{n(n+1)(2n+1)}{6}}
\end{equation*}
Sum of the cubes of the first n numbers:
\begin{equation*}
{\sum {n}^{{3}}=1^{{3}}+2^{{3}}+3^{{3}}+4^{{3}}\text{.}\text{.}\text{.}+n^{{3}}=\frac{n^{{2}}(n+1)^{{2}}}{4}}
\end{equation*}
Arithmetic Series
Say we have a geometric series as such:
$2+4+6+\dotso+96+98+100$
Each term is dependant off the previous one by adding a constant difference, say d, so that any term can be expressed as:
\begin{equation*}
{s_{{n}}=\left[a_{{1}}+(n-1)d\right]}
\end{equation*}
And the sum of n terms is:
\begin{equation*}
{\overset{{n}}{\underset{{n=1}}{\sum }}{a_{{i}}}=\frac{n}{2}(a_{{1}}+a_{{n}})}
\end{equation*}
An arithmetic series is always divergent.
Geometric Series
A Geometric series is any series that takes the form:
\begin{equation*}
{\overset{{\infty }}{\underset{{n=1}}{\sum }}{\normalsubformula{\text{ar}}^{{n-1}}}}
\end{equation*}
Each term within a geometric series can be expressed as:
\begin{equation*}
{s_{{n}}=\frac{a(1-r^{{n}})}{1-r}}
\end{equation*}
And the sum the infinite series is:
\begin{equation*}
{\overset{{n}}{\underset{{n=1}}{\sum }}{\normalsubformula{\text{ar}}^{{n-1}}}=\frac{a}{1-r}}
\end{equation*}
The geometric series will converge if ${|r|<1}$ and diverge on all other values.
P series and Harmonic Series
A P Series takes on the form:
\begin{equation*}
{\overset{{\infty }}{\underset{{n=1}}{\sum }}{\frac{1}{n^{{p}}}}}
\end{equation*}
When p=1 we get a Harmonic Series in the form:
\begin{equation*}
{\overset{{\infty }}{\underset{{n=1}}{\sum }}{\frac{1}{n}}}
\end{equation*}
Where each term can be expressed as:
\begin{equation*}
{s_{{n}}=\frac{1}{a+(n-1)d}}
\end{equation*}
And the sum the infinite series is:
\begin{equation*}
{\overset{{n}}{\underset{{n=1}}{\sum }}{\frac{1}{n}}=\frac{2}{n(2a+(n-1)d)}}
\end{equation*}
The P series is convergent when ${p>1}$ and divergent when ${p\le 1}$. Since the Harmonic series is a p series with p=1, it is always divergent.
Ratio Test
Suppose we have a series ${\sum {a_{{n}}}}$ where:
${L=\underset{{n\rightarrow \infty }}{{\text{lim}}}\left(\frac{a_{{n+1}}}{a_{{n}}}\right)}$ exists
Then,
If L$<$1 the series is absolutely convergent
If L$>$1 the series is divergent
If L=1 the series may be divergent or convergent.
Power Series
A power series is an infinite series in the form:
\begin{equation*}
{\overset{{\infty }}{\underset{{n\rightarrow 0}}{\sum
}}{a_{{n}}(x-c)^{{n}}}=a_{{o}}+a_{{1}}(x-c)^{{1}}+a_{{2}}(x-c)^{{2}}+\text{.}\text{.}\text{.}+a_{{n}}(x-c)^{{n}}}
\end{equation*}
Where ${a_{{n}}}$ corresponds to the coefficient of the ${n^{{\text{th}}}}$ term, c is a constant, and x varies around c. In many cases, the c is zero, in which the power series takes that form:
\begin{equation*}
{\overset{{\infty }}{\underset{{n\rightarrow 0}}{\sum
}}{a_{{n}}x^{{n}}}=a_{{o}}+a_{{1}}x^{{1}}+a_{{2}}x^{{2}}+\text{.}\text{.}\text{.}+a_{{n}}x^{{n}}}
\end{equation*}
Taylor and Maclaurin Series
If ${f(x)}$ is infinitely differentiable, then the series is defined as a Taylor Series:
\begin{equation*}
{\overset{{\infty }}{\underset{{n\rightarrow 0}}{\sum }}{\frac{f^{{(n)}}(c)}{n!}}(x-c)^{{n}}}
\end{equation*}
In the case that c=0, then the series becomes a Maclaurin series.
