2.7 Infinite series  (Page 3/3)

The Comparison Test is the most powerful theorem we have about infinite series of positive terms. Of course, most series do not consist entirely of positive terms, so that the Comparison Test is not enough.The next theorem is therefore of much importance.

Alternating series test

Suppose $\{a_1,a_2,a_3,...\}$ is an alternating sequence of real numbers; i.e., their signs alternate.Assume further that the sequence $\left\{\right|a_n\left|\right\}$ is nonincreasing with $0=lim|a_n|.$ Then the infinite series $\sum a_n$ converges.

Assume, without loss of generality, that the odd terms $a_{2n+1}$ of the sequence $\left\{a_n\right\}$ are positive and the even terms $a_{2n}$ are negative. We collect some facts about the partial sums $S_N=a_1+a_2+...+a_N$ of the infinite series $\sum a_n.$

1. Every even partial sum $S_{2N}$ is less than the following odd partial sum $S_{2N+1}=S_{2N}+a_{2N+1},$ And every odd partial sum $S_{2N+1}$ is greater than the following even partial sum $S_{2N+2}=S_{2N+1}+a_{2N+2}.$
2. Every even partial sum $S_{2N}$ is less than or equal to the next even partial sum $S_{2N+2}=S_{2N}+a_{2N+1}+a_{2N+2},$ implying that the sequence of even partial sums $\left\{S_{2N}\right\}$ is nondecreasing.
3. Every odd partial sum $S_{2N+1}$ is greater than or equal to the next odd partial sum $S_{2N+3}=S_{2N+1}+a_{2N+2}+a_{2N+3},$ implying that the sequence of odd partial sums $\left\{S_{2N+1}\right\}$ is nonincreasing.
4. Every odd partial sum $S_{2N+1}$ is bounded below by $S_2.$ For, $S_{2N+1}>S_{2N}\ge S_2.$ And, every even partial sum $S_{2N}$ is bounded above by $S_1.$ For, $S_{2N}<S_{2N+1}\le S_1.$
5. Therefore, the sequence $\left\{S_{2N}\right\}$ of even partial sums is nondecreasing and bounded above.That sequence must then have a limit, which we denote by $S_e.$ Similarly, the sequence $\left\{S_{2N+1}\right\}$ of odd partial sums is nonincreasingand bounded below. This sequence of partial sums also must have a limit, which we denote by $S_o.$

Now

showing that $S_e=S_o,$ and we denote this common limit by $S.$ Finally, given an $ϵ>0,$ there exists an $N_1$ so that $|S_{2N}-S|<ϵ$ if $2N\ge N_1,$ and there exists an $N_2$ so that $|S_{2N+1}-S|<ϵ$ if $2N+1\ge N_2.$ Therefore, if $N\ge max(N_1,N_2),$ then $|S_N-S|<ϵ,$ and this proves that the infinite series converges.

Test for irrationality

Let $x$ be a real number, and suppose that $\{p_N/q_N\}$ is a sequence of rational numbers for which $x=lim{p}_N/q_N$ and $x\ne p_N/q_N$ for any $N.$ If $lim{q}_N|x-p_N/q_N|=0,$ then $x$ is irrational.

We prove the contrapositive statement; i.e., if $x=p/q$ is a rational number, then $lim{q}_N|x-p_N/q_N|\ne 0.$ We have

Now the numerator $p{q}_N-q{p}_N$ is not 0 for any $N.$ For, if it were, then $x=p/q=p_N/q_N,$ which we have assumed not to be the case. Therefore, since $p{q}_N-q{p}_N$ is an integer, we have that

So,

and this clearly does not converge to 0.