Tail of Convergent Series tends to Zero

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Theorem

Let $\left\langle{a_n}\right\rangle_{n \geq 1}$ be a sequence of real numbers.

Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be a convergent series.

Let $N \in \N_{\ge 1}$ be a natural number.

Let $\displaystyle \sum_{n \mathop = N}^\infty a_n$ be the tail of the series $\displaystyle \sum_{n \mathop = 1}^\infty a_n$.


Then:

$\displaystyle \sum_{n \mathop = N}^\infty a_n$ is convergent
$\displaystyle \sum_{n \mathop = N}^\infty a_n \to 0$ as $N \to \infty$.

That is, the tail of a convergent series tends to zero.


Proof

Let $\left \langle {s_n} \right \rangle$ be the sequence of partial sums of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$.

Let $\left \langle {s'_n} \right \rangle$ be the sequence of partial sums of $\displaystyle \sum_{n \mathop = N}^\infty a_n$.


It will be shown that $\left \langle {s'_n} \right \rangle$ fulfils the Cauchy criterion.

That is:

$\forall \epsilon \in \R_{>0}: \exists N: \forall m, n > N: \left\vert{s'_n - s'_m}\right\vert < \epsilon$


Let $\epsilon \in \R_{>0}$ be a strictly positive real number.

As $\left \langle{s_n}\right \rangle$ is convergent, it conforms to the Cauchy criterion by Convergent Sequence is Cauchy Sequence.

Thus:

$\exists N: \forall m, n > N: \left\vert{s_n - s_m}\right\vert < \epsilon$

Now:

\(\displaystyle s_n\) \(=\) \(\displaystyle \sum_{k \mathop = 1}^n a_k\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^{N - 1} a_k + \sum_{k \mathop = N}^n a_k\) Indexed Summation over Adjacent Intervals
\(\displaystyle \) \(=\) \(\displaystyle s_{N - 1} + s'_n\)

and similarly:

$s_m = s_{N - 1} + s'_m$

Thus:

$s'_n = s_n - s_{N - 1}$

and:

$s'_m = s_m - s_{N - 1}$

So:

\(\displaystyle \left\vert{s_n - s_m}\right\vert\) \(<\) \(\displaystyle \epsilon\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{s_n - s_{N - 1} - s_m + s_{N - 1} }\right\vert\) \(<\) \(\displaystyle \epsilon\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{\left({s_n - s_{N - 1} }\right) - \left({s_m - s_{N - 1} }\right)}\right\vert\) \(<\) \(\displaystyle \epsilon\)
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{\left\vert{s_n - s_{N - 1} }\right\vert - \left\vert{s_m - s_{N - 1} }\right\vert}\right\vert\) \(<\) \(\displaystyle \epsilon\) Triangle Inequality
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{s'_n - s'_m}\right\vert\) \(<\) \(\displaystyle \epsilon\)

So $\displaystyle \sum_{n \mathop = N}^\infty a_n$ fulfils the Cauchy criterion.

By Convergent Sequence is Cauchy Sequence it follows that it is convergent.


Now it is shown that $\displaystyle \sum_{n \mathop = N}^\infty a_n \to 0$ as $N \to \infty$.

We have that $\left \langle {s_n} \right \rangle$ is convergent,

Let its limit be $l$.

Thus we have:

$\displaystyle l = \sum_{n \mathop = 1}^\infty a_n = s_{N - 1} + \sum_{n \mathop = N}^\infty a_n$

So:

$\displaystyle \sum_{n \mathop = N}^\infty a_n = l - s_{N - 1}$

But $s_{N - 1} \to l$ as $N - 1 \to \infty$.

The result follows.

$\blacksquare$


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