3.2 The kkth term test for series

In the rest of this chapter, we shall develop a systematic theory for determining whether series converge or diverge. In contrast with much of the above, we shall avoid as far as possible relying on explicit formulæ  for the partial sums which, in general, are hard or impossible to come by.

Our first result is a simple criterion which can be used to rule out convergence.

Lemma 3.18 (kkth term test).

Let (ak)kβˆˆβ„•(a_{k})_{k\in\mathbb{N}} be a sequence of real numbers. If the series βˆ‘k=1∞ak\sum_{k=1}^{\infty}a_{k} converges, then akβ†’0a_{k}\to 0 as kβ†’βˆžk\to\infty.

Proof.

Suppose the series βˆ‘k=1∞ak\sum_{k=1}^{\infty}a_{k} converges and let (sn)nβˆˆβ„•(s_{n})_{n\in\mathbb{N}} denote the sequence of partial sums. By definition, (sn)nβˆˆβ„•(s_{n})_{n\in\mathbb{N}} converges to some sβˆˆβ„s\in\mathbb{R} and so does the shifted sequence (sn+1)nβˆˆβ„•(s_{n+1})_{n\in\mathbb{N}}.22 2 This can be seen using either the subsequence test, or by arguing directly using the Ξ΅\varepsilon-NN definition. Since an+1:=sn+1βˆ’sna_{n+1}:=s_{n+1}-s_{n}, it follows from the limit laws for sequences that an+1β†’sβˆ’s=0a_{n+1}\to s-s=0 as nβ†’βˆžn\to\infty. The desired result now immediately follows. ∎

The contrapositive of the kkth term test can be a useful tool for showing that various series diverge. For instance, returning to the series βˆ‘k=1∞1\sum_{k=1}^{\infty}1 and βˆ‘k=1∞(βˆ’1)n\sum_{k=1}^{\infty}(-1)^{n} from ExampleΒ 3.6 and ExampleΒ 3.7, we can use the kkth term test to see that these diverge, since neither the sequence (1)kβˆˆβ„•(1)_{k\in\mathbb{N}} nor the sequence ((βˆ’1)k)kβˆˆβ„•((-1)^{k})_{k\in\mathbb{N}} converges to 0.

Warning 3.19.

It is not possible to use the kkth term test to show series converge, since the logical implication only runs in one direction (β€œif the series converges then its terms must converge to 0”). The reverse implication is false. To illustrate this, recall from ExampleΒ 2.34 that the sequence (k+1βˆ’k)kβˆˆβ„•\big{(}\sqrt{k+1}-\sqrt{k}\big{)}_{k\in\mathbb{N}} satisfies k+1βˆ’kβ†’0\sqrt{k+1}-\sqrt{k}\to 0 as kβ†’βˆžk\to\infty. However, from ExampleΒ 3.16 we know that the series βˆ‘k=1∞(k+1βˆ’k)\sum_{k=1}^{\infty}\big{(}\sqrt{k+1}-\sqrt{k}\big{)} formed from this sequence diverges.

Exercise 3.20.

Show that the following series diverge.

  1. (i)
    ​

    βˆ‘k=1∞(βˆ’1)kβ‹…kk+1\displaystyle\sum_{k=1}^{\infty}(-1)^{k}\cdot\frac{k}{k+1};

  2. (ii)
    ​

    βˆ‘k=1∞k23⁒k2βˆ’k\displaystyle\sum_{k=1}^{\infty}\frac{k^{2}}{3k^{2}-k};

  3. (iii)
    ​

    βˆ‘k=1∞3k\displaystyle\sum_{k=1}^{\infty}3^{k}.