Tail of Convergent Series

Theorem

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

Let $N \in \N^*$ be a natural number.

The expression $\displaystyle \sum_{n=N}^\infty a_n$ is known as the tail of the series $\displaystyle \sum_{n=1}^\infty a_n$.

Then:

• $\displaystyle \sum_{n=N}^\infty a_n$ is convergent;
• $\displaystyle \sum_{n=N}^\infty a_n \to 0$ 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 $\displaystyle \sum_{n=1}^\infty a_n$.

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

• We are going to show that $\left \langle s'_n \right \rangle$ fulfils the Cauchy criterion.

That is:

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

So, let $\epsilon > 0$.

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:

$\displaystyle s_n = \sum_{k=1}^n a_k = \sum_{k=1}^{N-1} a_k + \sum_{k=N}^n a_k = s_{N-1} + s'_n$ and similarly $s_m = s_{N-1} + s'_m$

Thus $s'_n = s_n - s_{N-1}, s'_m = s_m - s_{N-1}$.

So:

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

So $\displaystyle \sum_{n=N}^\infty a_n$ fulfils the Cauchy criterion, and by Convergent Sequence is Cauchy Sequence is convergent.

• Now we show that $\displaystyle \sum_{n=N}^\infty a_n \to 0$ as $N \to \infty$.

We know that $\left \langle s_n \right \rangle$ is convergent. Let its limit be $l$.

Thus we have $\displaystyle l = \sum_{n=1}^\infty a_n = s_{N-1} + \sum_{n=N}^\infty a_n$.

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

But $s_{N-1} \to l$ as $N-1 \to \infty$ and the result follows.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 6.11$