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 $s_N$.

$\blacksquare$