PlanetMath (more info)
 Math for the people, by the people. Sponsor PlanetMath
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
Revision difference : if $\sum_{k=1}^\infty a_k$ converges then $a_k\to 0$
Version 4 Version 3
\begin{thm} Suppose $a_1,a_2, \ldots$ is a sequence of real or complex numbers. \begin{thm} Suppose $a_1,a_2, \ldots$ is a sequence of real or complex numbers.
If the series If the infinite sum
$$ $$
\sum_{k=1}^\infty a_k \sum_{k=1}^\infty a_k
$$ $$
converges, then $\lim_{k\to \infty} a_k = 0$. converges, then $\lim_{k\to \infty} a_k = 0$.
\end{thm} \end{thm}
\begin{proof} Let $S\in \C$ be the value of the sum, and let $\varepsilon>0$ \begin{proof} Let $S\in \C$ be the value of the sum, and let $\varepsilon>0$
be arbitrary. Then there exists an $N\ge 1$ such that be arbitrary. Then there exists an $N\ge 1$ such that
$$ $$
| \sum_{k=1}^M a_k -S | < \frac{\varepsilon}{2} | \sum_{k=1}^M a_k -S | < \frac{\varepsilon}{2}
$$ $$
for all $M\ge N$. For $j\ge N$ we then have for all $M\ge N$. For $j\ge N$ we then have
\begin{eqnarray*} \begin{eqnarray*}
|a_{j+1}| &=& | \sum_{k=1}^{j+1} a_k -\sum_{k=1}^j a_k| \\ |a_{j+1}| &=& | \sum_{k=1}^{j+1} a_k -\sum_{k=1}^j a_k| \\
&\le & | \sum_{k=1}^{j+1} a_k -S | + |S - \sum_{k=1}^j a_k| \\ &\le & | \sum_{k=1}^{j+1} a_k -S | + |S - \sum_{k=1}^j a_k| \\
&<& \varepsilon, &<& \varepsilon,
\end{eqnarray*} \end{eqnarray*}
and the claim follows. and the claim follows.
\end{proof} \end{proof}