Such-that.com Series - survival kit

Definition of series

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.

Classic series

• Geometric Series

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 $$p\leqslant 1$$.

Convergence criteria

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.

Series with nonnegative terms

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 $$N$$, $$0\leqslant a_n\leqslant b_n$$.

• If $$\sum b_n$$ is convergent, then $$\sum a_n$$ is convergent.
• If $$\sum a_n$$ is divergent, then $$\sum b_n$$ is divergent.

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.

Absolute convergence and conditional convergence

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.