Convergent Series of Natural Numbers
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Theorem
Let $\sequence {a_n}_{n \mathop \in \N}$ be a sequence of natural numbers.
Then the following are equivalent:
$(1): \quad \ds \sum_{n \mathop = 1}^\infty a_n$ converges
$(2): \quad \exists N \in \N: \forall n \ge N: a_n = 0$
That is, $\ds \sum_{n \mathop = 1}^\infty a_n$ converges if and only if only finitely many of the $a_n$ are non-zero.
Proof
$(1) \implies (2)$:
Suppose that there is an infinite subsequence $\sequence {a_{n_k} }_{k \mathop \in \N}$ such that for each $k$, $a_{n_k} \ne 0$.
For $N \in \N$ let
- $\ds s_N = \sum_{n \mathop = 1}^N a_n$
To show that $s_N$ diverges it suffices to show that:
- $\forall M > 0: \exists N \in \N : \forall n > N : \size {s_n} > M$
Since for each $n$, $a_n \ge 0$, $s_N$ is a positive increasing sequence in $N$.
Therefore it suffices to show that:
- $\forall M > 0: \exists N \in \N: s_N > M$
Fix $M > 0$.
Let $k$ be any positive integer such that $n_k > M$.
Then we have:
\(\ds s_{n_k}\) | \(=\) | \(\ds \sum_{n \mathop = 1}^{n_k} a_n\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds \sum_{n \mathop = 1}^{n_k} 1\) | as the $a_n$ are positive and non-zero | |||||||||||
\(\ds \) | \(=\) | \(\ds n_k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds M\) | By the choice of $n_k$ |
Therefore the sequence $s_N$ diverges.
$(2) \implies (1)$:
Suppose there exists $N > 0$ such that $a_n = 0$ for all $n > N$.
Then we have, for all $L > N$:
- $\ds s_L = \sum_{n \mathop = 1}^L a_n = \sum_{n \mathop = 1}^N a_n = s_N$
In particular, for any $\epsilon > 0$ and all $L > N$:
- $\size {s_L - s_N} = 0 < \epsilon$
Therefore the sequence converges to $s_N$.
$\blacksquare$