Given a sequence \(u_n\), you can play with it and create a new sequence \(S_n\) defined by
\(S_1=u_1\)
\(S_2=u_1+u_2\)
\(S_3=u_1+u_2+u_3\)
More generally \(S_n=u_1+u_2+\cdots+u_n=\sum_{k=1}^{n}u_k\)
\(S_n\) is called the sequence of partial sums.
It is a sequence and any results about sequences can be used for \(S_n\).
The series \(\sum u_n\) or \(\sum_{k=0}^{\infty} u_n\) is the limit of \(S_n\) when \(n\) goes to \(\infty\).
Basically a series is a real number, the value of the limit when \(S_n\) is convergent.
We say the the series is convergent when the sequence \(S_n\) is convergent, has a finite limit. In this case the series is equal to the value of the limit.
The series is divergent when the limit of \(S_n\) is infinite or doesn't exist.
A geometric series with ratio \(r\) is \( \sum_{k=0}^{\infty}r^k\).
We can prove that its sequence of partial sum \(S_n=\sum_{k=0}^{\infty} r^k=\frac{1-r^{n+1}}{1-r}\) for \(r\neq 1\) and \(S_n=n+1\) for \(r=1\).
Multiply \(S_n\) by \((1-r)\), after simplifications, you will find that \(S_n(1-r)=1-r^{n+1}\)
By taking the limit as \(n\) goes to \(\infty\), we can conclude that
Theorem: The geometric series of ratio \(r\) is convergent for \(r\in(-1,1)\) and the series
$$\sum_{k=0}^{\infty}r^k=\frac{1}{1-r}.$$
The geometric series is divergent for \(r\notin(-1,1)\).
p-series
A p-series is a series in the form $$\sum_{n=1}^{\infty}\frac{1}{n^p}$$
P-test: the p-series \(\sum_{n=1}^{\infty} \frac{1}{n^p}\) is convergent for \(p>1\)
The p-series is divergent for \(p\leqslant 1\).
Theorem: If a series \(\sum u_n\) is convergent then \(\lim_{n\rightarrow\infty}u_n=0\)
Remark: The converse is FALSE. The series \(\sum\frac{1}{n}\) is divergent and \(\lim_{n\rightarrow\infty}\frac{1}{n}=0\).
The contrapositive is TRUE and is probably the most used form of the theorem: if \(\lim_{n\rightarrow \infty}u_n\neq 0\) then the series is divergent.
Example: \(\lim_{n\rightarrow\infty}\frac{n^2+\sin n}{3n^2+\sqrt{n}}=\frac{1}{3}\neq 0\) therefore the series $$\sum_{n=1}^{\infty}\frac{n^2+\sin n}{3n^2+\sqrt{n}}$$ is divergent.
Theorem (Comparison test, Inequality version) Let \(\sum a_n\) and \(\sum b_n\) series with non negative terms such that for all the index greater than some \(N\), \(0\leqslant a_n\leqslant b_n\).
Theorem (Comparison test, limit version) Let \(\sum a_n\) and \(\sum b_n\) series with positive terms such that \(\lim_{n\rightarrow\infty} \frac{a_n}{b_n}=c\neq 0\), then either \(\sum a_n\) and \(\sum b_n\) are both convergent or \(\sum a_n\) and \(\sum b_n\) are both divergent.
Theorem (Ratio Test) Let \(\sum a_n\) and \(\sum b_n\) series with positive terms such that \(\frac{a_{n+1}}{a_n}\leqslant \frac{b_{n+1}}{b_n}\) for any \(n\),
then
If \(\sum b_n\) is convergent, then \(\sum a_n\) is convergent.
If \(\sum a_n\) is divergent, then \(\sum b_n\) is divergent.
Corollary (Comparison to a geometric series) If for any \(n\), \(\frac{u_{n+1}}{u_n}<r<1\), then \(\sum u_n\) is convergent.
Corollary: If \(\lim_{n\rightarrow \infty}\frac{u_{n+1}}{u_n}=L\)
if \(L>1\), then the series with general term \(u_n\) is divergent,
if \( L<1\), the series with general terms \(u_n\) is convergent.
if \(L=1\), we cannot conclude.
Theorem (comparison with an integral)
Given a positive function \(f\) that is non increasing on a interval \([1,\infty)\).
Then the series \(\sum f(n)\) and \(\int_1^{\infty} f(t)d t\) are both convergent or they are both divergent.
Definition Given a series \(\sum u_n\), \(\sum u_n\) is absolutely convergent if \(\sum |u_n|\) is convergent.
Theorem If \(\sum u_n\) is absolutely convergent, then \(\sum u_n\) is convergent.
Definition If \(\sum u_n\) is convergent and not absolutely convergent, then \(\sum u_n\) is conditionally convergent.
Theorem (Alternating series) A series \(\sum u_n\) uch that its terms alternate between positive and negative, and such that \(|u_n|\) is decreasing toward 0 is convergent.