3.3 Limit laws for series

Since series are a particular kind of sequence, we can take any result from Chapter 2 about sequences and reformulate it as a result about series.

Addition and scalar multiplication of series

Here we convert the limit laws for addition and scalar multiplication of sequences into the language of series.

Theorem 3.21 (Limit laws for series).

Let (ak)k∈ℕ(a_{k})_{k\in\mathbb{N}} and (bk)k∈ℕ(b_{k})_{k\in\mathbb{N}} be sequences such that the series ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} and ∑k=1∞bk\sum_{k=1}^{\infty}b_{k} both converge.

  1. 1
    ​

    The series ∑k=1∞(ak+bk)\sum_{k=1}^{\infty}(a_{k}+b_{k}) converges and

    ∑k=1∞(ak+bk)=∑k=1∞ak+∑k=1∞bk.\sum_{k=1}^{\infty}(a_{k}+b_{k})=\sum_{k=1}^{\infty}a_{k}+\sum_{k=1}^{\infty}b% _{k}.
  2. 2
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    For λ∈ℝ\lambda\in\mathbb{R}, the series ∑k=1∞(λ⋅ak)\sum_{k=1}^{\infty}(\lambda\cdot a_{k}) converges and

    ∑k=1∞(λ⋅ak)=λ⁢∑k=1∞ak.\sum_{k=1}^{\infty}(\lambda\cdot a_{k})=\lambda\sum_{k=1}^{\infty}a_{k}.
Proof.

The proof follows by observing the partial sums satisfy

∑k=1n(ak+bk)=∑k=1nak+∑k=1nbkand∑k=1n(λ⋅ak)=λ⁢∑k=1nakfor all n∈ℕ\sum_{k=1}^{n}(a_{k}+b_{k})=\sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}b_{k}\qquad\text% {and}\qquad\sum_{k=1}^{n}(\lambda\cdot a_{k})=\lambda\sum_{k=1}^{n}a_{k}\qquad% \text{for all $n\in\mathbb{N}$}

and applying the limit laws for sequences from Theorem 2.38. We leave the details as an exercise (see Exercise 3.22 below). ∎

Exercise 3.22.

Prove Theorem 3.21. Your argument should combine the definition of a convergent series (in terms of the partial sums) and the limit laws for sequences.

Tails

For a given series ∑k=1∞ak\sum_{k=1}^{\infty}a_{k}, it can sometimes be useful to consider the tail ∑k=m+1∞ak\sum_{k=m+1}^{\infty}a_{k}, for some m∈ℕm\in\mathbb{N}. Such a tail is also a series, with its sequence of partial sums (tn)n∈ℕ(t_{n})_{n\in\mathbb{N}} given by

(3.4) (3.4) tn:=∑k=m+1m+nak=sm+n−sm,t_{n}:=\sum_{k=m+1}^{m+n}a_{k}=s_{m+n}-s_{m},

where sn:=∑k=1naks_{n}:=\sum_{k=1}^{n}a_{k} are the partial sums of the original series. Note that sms_{m} is a finite value that does not depend on nn.

If the original series ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges to ss, taking the limit as n→∞n\to\infty in (3.4) and using the limit laws for sequences, we have tn→s−smt_{n}\to s-s_{m} as n→∞n\to\infty, so the tail converges and the limits satisfy

∑k=m+1∞ak=∑k=1∞am+kand∑k=1∞ak=∑k=1mak+∑k=m+1∞ak.\sum_{k=m+1}^{\infty}a_{k}=\sum_{k=1}^{\infty}a_{m+k}\qquad\text{and}\qquad% \sum_{k=1}^{\infty}a_{k}=\sum_{k=1}^{m}a_{k}+\sum_{k=m+1}^{\infty}a_{k}.

Conversely, if some tail ∑k=m+1∞ak\sum_{k=m+1}^{\infty}a_{k} converges, then so too does the original series ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} and the above formulæ  hold (why?).

Exercise 3.23 (Tail test).

Let (ak)k∈ℕ(a_{k})_{k\in\mathbb{N}} be a sequence such that ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges. Show that the sequence of tails (∑k=m+1∞ak)m∈ℕ\big{(}\sum_{k=m+1}^{\infty}a_{k}\big{)}_{m\in\mathbb{N}} converges to 0.

Example 3.24.

For aa, r∈ℝr\in\mathbb{R} with |r|<1|r|<1, we have ∑k=m+1∞a⁢rk=a⁢rm+11−r\sum_{k=m+1}^{\infty}ar^{k}=\frac{ar^{m+1}}{1-r} for all m∈ℕ0m\in\mathbb{N}_{0}.

Proof.

Since the series ∑k=1∞a⁢rk\sum_{k=1}^{\infty}ar^{k} converges, we know the tail ∑k=m+1∞a⁢rk\sum_{k=m+1}^{\infty}ar^{k} converges and

∑k=m+1∞a⁢rk=∑k=1∞a⁢rm+k=rm⁢∑k=1∞a⁢rk,\sum_{k=m+1}^{\infty}ar^{k}=\sum_{k=1}^{\infty}ar^{m+k}=r^{m}\sum_{k=1}^{% \infty}ar^{k},

where the last identity follows by writing rm+k=rm⁢rkr^{m+k}=r^{m}r^{k} and applying the limit laws. The result now follows from the formula (3.2) for the limit of a geometric series. ∎

In some cases it is also convenient to start the summation from the index 0.

Exercise 3.25.

Use (3.2) to show that ∑k=0∞a⁢rk=a1−r\sum_{k=0}^{\infty}ar^{k}=\frac{a}{1-r} for r∈ℝr\in\mathbb{R} with |r|<1|r|<1.