if $\sum_{k=1}^\infty a_k$ converges then $a_k\to 0$ThenA_kto0IfSum_k1inftyA_kConverges2005-02-06 10:11:532005-11-18 12:42:05Theoremseriesdivergence test% this is the default PlanetMath preamble. as your knowledge
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\newtheorem{thm}{Theorem}
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\newtheorem{cor}{Corollary}\begin{thm} Suppose $a_1,a_2, \ldots$ is a sequence of real or complex numbers.
If the series
$$
\sum_{k=1}^\infty a_k
$$
converges, then $\lim_{k\to \infty} a_k = 0$.
\end{thm}
\subsubsection*{Remarks}
\begin{enumerate}
\item
The harmonic series $\sum_{k=1}^\infty 1/k$ shows that the
implication can not be reversed.
\item This result can be used as a first test for convergence of a series
$\sum_{k=1}^\infty a_k$. If $a_k$ does not converge to $0$, then
$\sum_{k=1}^\infty a_k$ does not converge either.
\end{enumerate}
\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
$$
| \sum_{k=1}^M a_k -S | < \frac{\varepsilon}{2}
$$
for all $M\ge N$. For $j\ge N$ we then have
\begin{eqnarray*}
|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| \\
&<& \varepsilon,
\end{eqnarray*}
and the claim follows.
\end{proof}