PX153 - F2 - convergence of a series

• in general, a series is just a sum of terms:

convergence

• some infinite series converge to a finite value, and some diverge
• eg:
  • $S_1=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}$ : converges
  • $S_2=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}\to \mathrm{\infty }$ : diverges
• a series with $N$ terms, $S_N=\sum _{n=1}^N{a}_n$, is convergent if $\underset{N\to \mathrm{\infty }}{lim}{S}_N$ exists and is finite
  • else, it is divergent
• sometimes we can determine whether a series will converge by algebraic manipulation:

- multiply both sides by $x$:
$$xS_{N} = \sum\limits_{n=1}^{N+1} x^{n} = S_{N} - 1 + x^{N+1}$$
$$(x-1)S_{N} = x^{N+1} - 1$$
$$\therefore S_{N}= \frac{x^{N+1}-1}{1-x}$$
- for large $N$, this gives $S_N\to \frac{1}{1-x}$, provided $|x|<1$. hence, this power series converges if $-1<x<1$. this is the interval of convergence
- the radius of convergence is half of the interval: $R=1$, that is $|x|<1$
- this is an example of a geometric series. more generally:
$$S_{N} = \sum\limits_{n=0}^{N} ar^{n}$$
- following the same argument, this converges for $|r|<1,limN\to \mathrm{\infty }{r}^N=0$ and the sum tends to $S=\frac{a}{1-r}$

tests for convergence

• $f(x)$ must be continuous and differentiable

preliminary test

• simple, but not conclusive
• a series, $S=\sum _{n=1}^{\mathrm{\infty }}{a}_n$, cannot converge unless $\underset{n\to \mathrm{\infty }}{lim}{a}_n=0$
• this is necessary, but insufficient condition for convergence
• eg: $S=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}$ diverges despite $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$

comparison test

• if $S_1=\sum _{n-1}^{\mathrm{\infty }}{a}_n$ is a series of positive terms, and $S_2=\sum _{n=1}^{\mathrm{\infty }}{b}_n$ is also a series of positive terms, then $S_1$ converges if $a_n\le b_n$ for large $n$
• conversely, if $a_n\ge b_n$ for large $n$, and $S_2$ diverges, then $S_1$ diverges
• eg: does $S_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)}$ converge?
  • since $\frac{1}{n(n+1)}<\frac{1}{n^2}$ for all $n\ge 1$, by the comparison test, $S_1$ converges

the ratio test (d'alembert's method)

• the ratio test states that for $S_1=\sum _{n-1}^{\mathrm{\infty }}{a}_n$ ; $\underset{n\to \mathrm{\infty }}{lim}|\frac{a_{n+1}}{a_n}|=k$

  • if $k<1$: converges
  • if $k>1$: diverges
  • if $k=1$: inconclusive

• proof:

  • consider $S=\sum _{n-1}^{\mathrm{\infty }}{a}_n$
  • such that $\underset{n\to \mathrm{\infty }}{lim}|\frac{a_{n+1}}{a_n}|=k<1$
  • as $k<1$, for the $N^{th}$ term, and beyond, in the series, there is a value of $r$, where $k<r<1$ and $\frac{a_{n+1}}{a_n}<r$ for all $n>N$
  • the terms in the series, $a_N$, $a_{N+1}$, $a_{N+2}$, $...$, are all less than $r{a}_N+r^2{a}_N+r^3{a}_N+...$
  • this is a geometric series with $r<1$, so, it converges.
  • by the comparison test, $S$ converges as well

• eg: does $S=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n{2}^n}$ converge?

  • preliminary test:
  $$\underset{n\to \mathrm{\infty }}{lim}=0$$

  - passed

  • ratio test:
  $$\underset{n\to \mathrm{\infty }}{lim}|\frac{1}{(n+1)2^{n+1}}n{2}^n|=\underset{n\to \mathrm{\infty }}{lim}|\frac{n}{2(n+1)}|=\underset{n\to \mathrm{\infty }}{lim}|\frac{1}{2}(1+\frac{1}{n})|$$$$lim=\frac{1}{2}=k$$

  - since $k<1$, $S$ converges

alternating series test

• if $S=\sum _{n-1}^{\mathrm{\infty }}{a}_n$ is a series with terms alternate in sign, and if the terms are continuously diminishing, with $\underset{n\to \mathrm{\infty }}{lim}=0$, then the series will converge