5.3 Series of Positive & Negative Terms

Up to this point we have only looked at methods for analysing the behaviour of series generated by sequences which only contain positive terms. This is very limiting and we will now expand the scope to general sequences.

Definition 5.4.

A series $\sum_{n=1}^{\infty}x_{n}$ is called absolutely convergent if $\sum_{n=1}^{\infty}\lvert x_{n}\rvert$ converges.

Theorem 5.5

Every absolutely convergent series is also convergent.

Proof.

Consider an absolutely convergent series $\sum_{n=1}^{\infty}x_{n}$ . Note that we have

x_{n}+\lvert x_{n}\rvert\leq 2\lvert x_{n}\rvert,\ \forall n\in\mathbb{N}.

(5.23)

Thus, by the comparison test (comparison with $\sum_{n=1}^{\infty}\lvert x_{n}\rvert$ ), the series $\sum_{n=1}^{\infty}(x_{n}+\lvert x_{n}\rvert)$ is convergent. Now, we can write

\sum_{n=1}^{\infty}x_{n}=\sum_{n=1}^{\infty}(x_{n}+\lvert x_{n}\rvert)-\sum_{n% =1}^{\infty}\lvert x_{n}\rvert,

(5.24)

and since $\sum_{n=1}^{\infty}x_{n}$ is the different of two convergent series, it is also convergent (follows from theorem 4.4). ∎

Example 5.6.

Show that the series $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}}$ is convergent.
Note that since $\left\lvert\frac{(-1)^{n}}{n^{2}}\right\rvert=\frac{1}{n^{2}}$ , by comparison with the series $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ the series $\sum_{n=1}^{\infty}\left\lvert\frac{(-1)^{2}}{n^{2}}\right\rvert$ is convergent. Hence the series $\sum_{n=1}^{\infty}\frac{(-1)^{2}}{n^{2}}$ is absolutely convergent and therefore convergent.

We will now look at two more convergence tests which will be useful.

5.3.1 The Root Test

Theorem 5.6 (The Root Test)

Let $\sum_{n=1}^{\infty}x_{n}$ be a series.

(i)

If $\exists N\in\mathbb{N}$ and $r<1$ such that for all $n\geq N,\lvert x_{n}\rvert^{\frac{1}{n}}\leq r$ , then the series converges.
(ii)

If $\exists N\in\mathbb{N}$ and $r>1$ such that for all $n\geq N,\lvert x_{n}\rvert^{\frac{1}{n}}\geq r$ , then the series diverges.

Proof.

(i)

For $n\geq N$ we have $\lvert x_{n}\rvert^{\frac{1}{n}}\leq r\implies\lvert x_{n}\rvert\leq r^{n}$ . Let $a_{n}=r^{n}$ , then $(a_{n})_{n}$ generates a convergent geometric series so by the comparison test the series $\sum_{n=1}^{\infty}\lvert x_{n}\rvert$ converges. Thus the series $\sum_{n=1}^{\infty}x_{n}$ is absolutely converges and therefore converges.
(ii)

For all $n\geq N$ we have $\lvert x_{n}\rvert^{\frac{1}{n}}\geq r\implies\lvert x_{n}\rvert\geq r^{n}$ , so $x_{n}$ must be a divergent sequence ( $\pm\infty$ ). In particular, $x_{n}$ does not converge to 0 as $n\to\infty$ , so by the contrapositive of theorem 5.1, $\sum_{n=1}^{\infty}x_{n}$ diverges.

∎

As with the ratio test, the root test says is that if $\lim_{n\to\infty}\lvert x_{n}\rvert^{\frac{1}{n}}$ exists and is less than 1, then the series converges, if it is greater than 1 then the series diverges and if it is equal to 1 then the root test gives us no information.

5.3.2 The Alternating Series Test

Theorem 5.7 (Alternating Series Test)

Let $(x_{n})_{n}$ be a positive decreasing sequence with limit 0 ( $\lim_{n\to\infty}x_{n}=0$ ). Then

\sum_{n=1}^{\infty}(-1)^{n}x_{n}

(5.25)

is a convergent series.

Proof.

∎

Example 5.7.

The sequence given by $x_{n}=\frac{1}{n}$ is positive and decreasing with limit 0. Thus by the alternating series test, $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}$ is convergent. Note that it is not absolutely convergent since $\left\lvert\frac{(-1)^{n}}{n}\right\rvert=\frac{1}{n}$ generates the harmonic series which diverges.

Theorem 5.8

These are some handy facts to keep in mind when using the root test.

(i)

$\lim_{n\to\infty}n^{\frac{1}{n}}=1$ .
(ii)

If $r>0$ , then $\lim_{n\to\infty}r^{\frac{1}{n}}=1$ .
(iii)

Suppose $(a_{n})_{n}$ is a positive sequence and $\frac{a_{n+1}}{a_{n}}\to r$ , then $a_{n}^{\frac{1}{n}}\to r$ .

Proof.

∎