Remainder (Tail) of a Series

The n-th remainder, or tail, of a series is the series obtained by discarding the first n terms of the original series. $$ R_n = \sum_{k=n+1}^{\infty} a_k \:\:\:\:\:\: n \in \mathbb{N} $$

A practical example

Consider the series

$$ \sum_{k=1}^{\infty} \frac{1}{k(k+1)} $$

The first partial sums are

$$ a_1 = \frac{1}{2} \\ a_2 = \frac{1}{2} + \frac{1}{6} \\ a_3 = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} \\ a_4 = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} \\ a_5 = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} \\ \vdots $$

The remainder R₃ is the tail of the series starting from k=4.

$$ R_n = \sum_{k=n+1}^{\infty} \frac{1}{k(k+1)} $$

$$ R_3 = \sum_{k=4}^{\infty} \frac{1}{k(k+1)} $$

The first terms of the tail are:

$$ b_1=0 \\ b_2=0 \\ b_3=0 \\ b_4 = \frac{1}{20} \\ b_5 = \frac{1}{20} + \frac{1}{30} \\ b_6 = \frac{1}{20} + \frac{1}{30} + \frac{1}{42} \\ \vdots $$

From a graphical perspective

Convergence of the remainder (tail)

If the series $$ \sum_{k=1}^{\infty} a_k $$ converges, then its remainder (tail) $$ R_n = \sum_{k=n+1}^{\infty} a_k $$ also converges.

Proof

Let $$ \sum_{k=1}^{\infty} a_k $$ be a convergent series, and define its n-th remainder as $$ R_n = \sum_{k=n+1}^{\infty} b_k $$

For every m>n, the terms of the two series coincide.

$$ a_m = b_m \:\:\: \forall m>n $$

Explanation. The remainder series R_n has its first n terms equal to zero. For k>n, its terms coincide with those of the original series. Hence, b_k=0 for k=1,\dots,n and b_k=a_k for k>n.

Therefore, for every m>n, the difference between corresponding terms is zero.

$$ b_m - a_m = 0 $$

Consequently, for every m>n, the partial sums of the difference series satisfy

$$ \sum_{k=1}^{m} (b_k - a_k) = \sum_{k=1}^{n} (b_k - a_k) $$

In other words, the sequence of partial sums of the series $$ \sum (b_k - a_k) $$ stabilizes for m>n.

Explanation. For k=1,\dots,n we have b_k=0, hence b_k-a_k=-a_k. For k>n, since a_k=b_k, we have b_k-a_k=0. Thus, adding further terms does not change the partial sum.

It follows that the series $$ \sum_{k=1}^{\infty} (b_k - a_k) $$ converges.

We now have two convergent series

$$ \sum_{k=1}^{\infty} a_k $$

$$ \sum_{k=1}^{\infty} (b_k - a_k) $$

By standard results on convergent series, their sum is also convergent and equals the series formed by summing the corresponding terms:

$$ \sum_{k=1}^{\infty} a_k + \sum_{k=1}^{\infty} (b_k - a_k) $$

$$ \sum_{k=1}^{\infty} \big(a_k + b_k - a_k \big) $$

$$ \sum_{k=1}^{\infty} b_k $$

This shows that the remainder (tail) R_n is convergent.

Example

Returning to the initial example

$$ \sum_{k=1}^{\infty} \frac{1}{k(k+1)} $$

$$ R_3 = \sum_{k=4}^{\infty} \frac{1}{k(k+1)} $$

Both series converge

$$ \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k(k+1)} = 1 $$

$$ \lim_{n \rightarrow \infty} \sum_{k=4}^{n} \frac{1}{k(k+1)} = 0.25 $$

Graphical representation

Corollary

If the series $$ \sum_{k=1}^{\infty} a_k $$ converges, then $$ \lim_{n \rightarrow \infty} R_n = 0 $$ where $$ R_n = \sum_{k=n+1}^{\infty} a_k $$

Proof

Let $$ \sum_{k=1}^{\infty} a_k $$ be a convergent series, and let $$ R_n = \sum_{k=n+1}^{\infty} b_k $$ be its remainder.

For every m>n, we have a_k=b_k for k>n, hence

$$ \sum_{k=1}^m (a_k - b_k) = \sum_{k=1}^n a_k $$

Explanation. For k=1,\dots,n we have b_k=0, so a_k-b_k=a_k. For k>n, a_k=b_k, so the difference is zero. Thus, the partial sums remain constant beyond n.

Taking the limit as m→∞

$$ \sum_{k=1}^{\infty} (a_k - b_k) = \sum_{k=1}^n a_k $$

Hence

$$ \sum_{k=1}^{\infty} a_k - \sum_{k=1}^{\infty} b_k = \sum_{k=1}^n a_k $$

Since $$ \sum b_k = R_n $$

$$ \left( \sum_{k=1}^{\infty} a_k \right) - R_n = \sum_{k=1}^n a_k $$

Taking the limit as n→∞

$$ \sum_{k=1}^{\infty} a_k - \lim_{n \rightarrow \infty} R_n = \sum_{k=1}^{\infty} a_k $$

Therefore

$$ \lim_{n \rightarrow \infty} R_n = 0 $$

This completes the proof.

Example

Returning to the initial example

$$ \lim_{n \rightarrow \infty} R_0 = \sum_{k=1}^{\infty} \frac{1}{k(k+1)} = 1 $$

$$ \lim_{n \rightarrow \infty} R_1 = \sum_{k=2}^{\infty} \frac{1}{k(k+1)} = 0.5 $$

$$ \lim_{n \rightarrow \infty} R_2 = \sum_{k=3}^{\infty} \frac{1}{k(k+1)} \approx 0.33 $$

$$ \lim_{n \rightarrow \infty} R_3 = \sum_{k=4}^{\infty} \frac{1}{k(k+1)} = 0.25 $$

As more initial terms are removed, the remainder R_n approaches zero.

$$ \lim_{n \rightarrow \infty} R_n = 0 $$

Graphical representation

And so on.