A series converges if its partial sums form a Cauchy sequence. As a consequence, for a convergent series, the underlying sequence necessarily needs to converge to zero.

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If a series converges can be checked with different tests. |

**Theorem 1** (Cauchy Criterion). *A series \(\displaystyle\sum_{k=1}^\infty a_k\) in
\(\mathbb{R}\) is convergent if and
only if for all \(\varepsilon>0\),
there exists some \(N\) such that for
all \(n\geq m\geq N\) holds \[\left|\sum_{k=m}^n
a_k\right|<\varepsilon.\]*

*Proof:* By the theorems on completeness and
convergence of cauchy sequences, a series converges if and only if the
sequence \((s_{n})_{n\in\mathbb{N}}\)
of partial sums is a Cauchy sequence.

On the other hand, for \(n\geq m\), we
have \[\left|s_n-s_{m-1}\right|=\left|\sum_{k=m}^n
a_k\right|.\] Therefore, the Cauchy criterion is really
equivalent to the fact that \((s_{n})_{n\in\mathbb{N}}\) is a Cauchy
sequence in \(\mathbb{R}\).\(\Box\)

As a corollary, we can formulate the following criterion.

**Theorem 2** (Necessary criterion for convergence of
series). *Let \[\sum_{k=1}^\infty
a_k\] be a convergent series in \(\mathbb{R}\). Then \((a_n)_{n \in \mathbb{N}}\) is convergent
with \[\lim_{n\to\infty}a_n=0.\]*

*Proof:* Since the series converges, the Cauchy
criterion implies that for all \(\varepsilon>0\), there exists some \(N\) such that for all \(n\geq m\geq N\) holds \[\left|\sum_{k=m}^n
a_k\right|<\varepsilon.\] Now considering the special case
\(n=m\), we have that for all \(n\geq N\) holds \[|a_n|<\varepsilon.\] However, this is
nothing but convergence of \((a_n)_{n\in\mathbb{N}}\) to zero. \(\Box\)

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