40-00-Ifsumk1inftyAkConvergesThenAkto0.tex

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\pmtitle{if $\sum_{k=1}^\infty a_k$ converges then $a_k\to 0$}
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\pmtype{Theorem}
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\pmsynonym{necessary condition of convergence}{Ifsumk1inftyAkConvergesThenAkto0}
%\pmkeywords{divergence test}
\pmrelated{DeterminingSeriesConvergence}
\pmrelated{CompleteUltrametricField}
\pmrelated{ConvergenceConditionOfInfiniteProduct}
\pmrelated{LambertSeries}
\pmrelated{AbsoluteConvergenceOfIntegralAndBoundednessOfDerivative}
\pmrelated{ConvergentSeriesWhereNotOnlyA_nButAlsoNa_nTendsTo0}

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\begin{document}
\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}
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