# Convergent Series of Natural Numbers

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## Theorem

Let $\left({a_n}\right)_{n \in \N}$ be a sequence of natural numbers.

Then the following are equivalent:

$(1): \quad \displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges

$(2): \quad \exists N \in \N: \forall n \ge N: a_n = 0$

That is, $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ 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 $\left({ a_{n_k} }\right)_{k \in \N}$ such that for each $k$, $a_{n_k} \neq 0$.

For $N \in \N$ let

$\displaystyle s_N = \sum_{n \mathop = 1}^N a_n$

To show that $s_N$ diverges it suffices to show that:

$\forall M > 0\ \exists N \in \N : \forall n > N : \left\vert{ s_n }\right\vert > M$

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:

 $\displaystyle s_{n_k}$ $=$ $\displaystyle \sum_{n \mathop = 1}^{n_k} a_n$ $\displaystyle$ $\ge$ $\displaystyle \sum_{n \mathop = 1}^{n_k} 1$ as the $a_n$ are positive and non-zero $\displaystyle$ $=$ $\displaystyle n_k$ $\displaystyle$ $=$ $\displaystyle M$ 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 $L > N$:

$\displaystyle s_L = \sum_{n \mathop = 1}^L a_n = \sum_{n \mathop = 1}^N a_n = s_N$

In particular, for any $\epsilon > 0$ and all $L > N$:

$\left\vert{s_L - s_N}\right\vert = 0 < \epsilon$

Therefore the sequence converges to $s_N$.

$\blacksquare$