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.
$\blacksquare$
Also see
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 6.11$