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More on Series¶

Check whether you recall the following:

What is $\limsup x_n$ for a sequence $(x_n)$?
What means by a series $\sum x_n$ converges? converges absolutely?
What is ratio test?

Proposition 2.6.1. (Root Test)

Let $\sum x_n$ be a series and let $L:=\limsup |x_n|^{1/n}$ (note that $L$ is either a real number $\ge 0$ or $+\infty$.)

Then

If $L < 1$, then $\sum x_n$ converges absolutely.
If $L > 1$, then $\sum x_n$ diverges.

See this page for a proof.

When $L = 1$, the test is inconclusive. For instance, $\sum 1/n$ diverges and $\sum 1/n^2$ converges.

But both $|1/n|^{1/n}$ and $|1/n^2|^{1/n}$ converges to $1$ as $n \to \infty$ because $n^{1/n} \to 1$.

Theorem. Let $(x_n)$ be a sequence of nonzero real numbers. Then

$$ \liminf \left|\frac{x_{n+1}}{x_n}\right| \le \liminf |x_n|^{1/n} \le \limsup |x_n|^{1/n} \le \limsup \left|\frac{x_{n+1}}{x_n}\right|. $$

Proof. Let us just proof the right-most inequality.

Let $L := \limsup\left|\frac{x_{n+1}}{x_n}\right|$ and $s := \limsup |x_n|^{1/n}$. The inequality is obvious when $L = +\infty$, so assume $L$ is finite.

To show that $s \le L$, it suffices to show that $s \le L'$ for any $L' > L$.

Pick any $L'$ bigger than $L = \lim_k \sup\left\{|x_{n+1}/x_n| \colon n \ge k \right\}$. Then there exists $N$ such that

$$ \sup \left\{ \left|\frac{x_{n+1}}{x_n}\right| \colon n \ge N \right\} < L'. $$

Thus, for $n \ge N$, $|x_{n+1}|/|x_n| < L'$. And so, for any $k \ge 1$ $$ \left| \frac{x_{N+k}}{x_N}\right| < (L')^k. $$

In other words (let $n = N+k$), for all $n > N$, $|x_n| < (L')^{n-N}|x_N|$ and so

$$ |x_n|^{1/n} \lt c_n:=(L')^{1-N/n}|x_N|^{1/n}. $$

Since $N$ and $L'$ are fixed, $c_n \to L' \cdot 1=L'$. In particular, $(c_n)$ and hence $(|x_n|^{1/n})$ is a bounded sequence. Therefore, (by the BW theorem) there exists a subsequence $(|x_{n_k}|^{1/n_k})$ of $(|x_n|^{1/n})$ that converges to $s$. And so we have

$$ s = \lim_k |x_{n_k}|^{1/n_k} \le \lim c_{n_k} = L'.$$

This is what needed to be shown.

Corollary The Root Test is stronger than the Ratio Test.

Proof Whenever $\lim |x_{n+1}/x_n|$ exists, $\limsup |x_n|^{1/n}$ exists and is the same value.

In fact, the Root Test is strictly stronger than the Ratio Test as the following example illustrates.

Example (Taken from K.A. Ross's Elementary Analysis)

Consider the series

$$ \sum_{n=0}^{\infty} 2^{(-1)^n-n} = 2+ \frac{1}{4} + \frac{1}{2} + \frac{1}{16} + \frac{1}{8}+ \frac{1}{64} +\cdots$$

Since $x_n = 2^{(-1)^n-n} \le \frac{1}{2^{n-1}}$ for all $n$, the series is convergent by the comparision test. However, since $x_{n+1}/x_n = 1/8$ when $n$ is even and $=2$ when $n$ is odd. Thus,

$$ \frac{1}{8} = \liminf \left|\frac{x_{n+1}}{x_n}\right| < 1 < \limsup\left|\frac{x_{n+1}}{x_n}\right| = 2.$$ So the Ratio Test gives no information.

However, $|x_n|^{1/n} = 2^{1/n-1}$ for even n and $|x_n|^{1/n} = 2^{-1/n-1}$ for odd $n$. Since $\lim 2^{1/n} = \lim 2^{-1/n} =1$, $\lim |x_n|^{1/n} = 1/2 < 1$. Therefore, the series converges by the root test.

Proposition 2.6.2 (Alternating series test).

Let $(x_n)$ be a monotonic decreasing null sequence. Then the series $\sum_{n=1}^{\infty} (-1)^n x_n$ coverges.

See this page for a proof.

Example. An application of the alternating series test (AST) implies

$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^p}$$

converges for arbitrary small $p > 0$, but it does not converges absolutely when $p \le 1$.

Theorem 2.6.5 (Mertens' Theorem) Suppose $\sum a_n$ and $\sum b_n$ converge to $A$ and $B$, respectively and at least one of the series converges absolutely.

Then the series $\sum c_n$, where $$ c_n =\sum_{i+j=n} a_ib_j $$ converges to $AB$.

See the textbook for a proof.

The series $\sum c_n$ is called the Cauchy product of $\sum a_n$ and $\sum b_n$.

Example 2.6.6 The Cauchy product of two conditionally convergent series may be divergent.

For instance, take both $\sum a_n$ and $\sum b_n$ to be the series $\ds \sum_{n\ge 1} \frac{(-1)^{n+1}}{\sqrt{n}}$.

Then for any $n \ge 1$, \begin{align*} |c_n| &= \left|\sum_{i+j = n} a_ib_j\right| = \left|\sum_{i+j=n} \frac{(-1)^{i+j}}{\sqrt{ij}}\right| \\ &=\left| (-1)^n \sum_{i+j=n} \frac{1}{\sqrt{ij}} \right| = \sum_{i+j=n} \frac{1}{\sqrt{ij}} \end{align*}

For $i,j \ge 1$ and $i+j =n$, $ij \lt i^2 + 2ij + j^2 = n^2$ so

\begin{align*} \sum_{i+j=n} \frac{1}{\sqrt{ij}} &\gt \sum_{i+j=n} \frac{1}{n} = \frac{n-1}{n} \to 1. \end{align*}

In particular, $c_n \not\to 0$ therefore, $\sum c_n$ diverges.