The algebra of convergent series.

Theorem 9.2.1   Let $ U=\underset{k=0}{\overset{+\infty }{\sum}} u_n$ and $ V=\underset{k=0}{\overset{+\infty }{\sum}} v_n$ be two convergent series. Then the series $ \underset{k=0}{\overset{+\infty }{\sum}} (u_n+v_n)$ is convergent and its sum is equal to $ U+V$ .

\begin{figure}\mbox{\subfigure{\epsfig{file=error.eps, height=1cm}} \qquad
The converse is not true!}

Theorem 9.2.2   Let $ U=\underset{k=0}{\overset{+\infty }{\sum}} u_n$ be a convergent series and $ \alpha \in \mathbb{R}$ . Then the series $ \underset{k=0}{\overset{+\infty }{\sum}} \alpha u_n$ is convergent and its sum is equal to $ \alpha U$ .

\begin{figure}\mbox{\subfigure{\epsfig{file=error.eps, height=1cm}} \qquad
The converse is true only for $\alpha \neq 0$.}

Together these two theorems mean that the set of convergent series of real numbers is a real vector space.

Remark 9.2.3   Adding or deleting terms from a series does not change the convergence/divergence. For a convergent series, it changes the value of the sum.

Example 9.2.4       

$\displaystyle \underset{k=2}{\overset{+\infty }{\sum}} \frac {1}{3^n} = \unders...
...\infty }{\sum}} \frac {1}{3^n} - 1 - \frac 13 = \frac 32 - \frac 43 = \frac 16.$    

Noah Dana-Picard 2007-12-28