# Tail of Convergent Series tends to Zero

## Theorem

Let $\sequence {a_n}_{n \mathop \ge 1}$ be a sequence of real numbers.

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

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

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

Then:

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

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

## Proof

Let $\sequence {s_n}$ be the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$.

Let $\sequence {s'_n}$ be the sequence of partial sums of $\ds \sum_{n \mathop = N}^\infty a_n$.

It will be shown that $\sequence {s'_n}$ fulfils the Cauchy criterion.

That is:

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

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

As $\sequence {s_n}$ is convergent, it conforms to the Cauchy criterion by Convergent Sequence is Cauchy Sequence.

Thus:

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

Now:

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

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:

 $\ds \size {s_n - s_m}$ $<$ $\ds \epsilon$ $\ds \leadsto \ \$ $\ds \size {s_n - s_{N - 1} - s_m + s_{N - 1} }$ $<$ $\ds \epsilon$ $\ds \leadsto \ \$ $\ds \size {\paren {s_n - s_{N - 1} } - \paren {s_m - s_{N - 1} } }$ $<$ $\ds \epsilon$ $\ds \leadsto \ \$ $\ds \size {\size {s_n - s_{N - 1} } - \size {s_m - s_{N - 1} } }$ $<$ $\ds \epsilon$ Triangle Inequality $\ds \leadsto \ \$ $\ds \size {s'_n - s'_m}$ $<$ $\ds \epsilon$

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

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

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

We have that $\sequence {s_n}$ is convergent,

Let its limit be $l$.

Thus we have:

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

So:

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

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

The result follows.

$\blacksquare$