Passing Absolute Convergence Down to Similar Series of Real Numb.

Passing Absolute Convergence Down to Similar Series of Real Numbers

Recall from the Absolute and Conditional Convergence of Series of Real Numbers page that a series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is said to be absolutely convergent if $\displaystyle{\sum_{n=1}^{\infty} \mid a_n \mid}$ converges. We also proved that every absolutely convergent series of real numbers is absolutely convergent.

We will see that if $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is an absolutely convergent series, then many other similar series that can be "constructed" from $\displaystyle{\sum_{n=1}^{\infty} a_n}$ are also absolutely convergent.

 Theorem 1: Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ be an absolutely convergent series. Then $\displaystyle{\sum_{n=1}^{\infty} a_n^2}$ is an absolutely convergent series.

We will give two proofs of Theorem 1.

• Proof 1): It suffices to show that $\displaystyle{\sum_{n=1}^{\infty} a_n^2}$ converges since a convergent series of nonnegative terms is always absolutely convergent. Let $(s_n)_{n=1}^{\infty}$ be the sequence of partial sums for this series. Then for each $n \in \mathbb{N}$ we have that:
(1)
\begin{align} \quad s_n = a_1^2 + a_2^2 + ... + a_n^2 \leq (\mid a_1 \mid + \mid a_2 \mid + ... + \mid a_n \mid)^2 \quad (*) \end{align}
• Now, let $(s_n')_{n=1}^{\infty}$ denote the sequence of partial sums to the series $\displaystyle{\sum_{n=1}^{\infty} \mid a_n \mid}$ which converges by the absolutely convergence of $\displaystyle{\sum_{n=1}^{\infty} a_n}$. Then we have that $\lim_{n \to \infty} s_n' = S$ for some $S \in \mathbb{R}$, $S > 0$. Moreover, by the properties of limits of convergent sequences we see that $\lim_{n \to \infty} s_n'^2 = S^2$ and from $(*)$ this means that $s_n \leq S^2$ for all $n \in \mathbb{N}$. But $(s_n)_{n=1}^{\infty}$ is an increasing sequence that is bounded above (by $S^2$), and so $(s_n)_{n=1}^{\infty}$ converges which shows that $\displaystyle{\sum_{n=1}^{\infty} a_n^2}$ converges absolutely. $\blacksquare$
• Proof 2) Suppose that $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges absolutely. Then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges, and $\lim_{n \to \infty} a_n = 0$. So for $\epsilon = 1 > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:
(2)
\begin{align} 0 \leq \quad \mid a_n \mid < 1 \quad \Leftrightarrow -1 < a_n < 1 \end{align}
• Notice that since for all $n \geq N$ we have that $-1 < a_n < 1$ that $a_n^2 < \mid a_n \mid$. So then $\displaystyle{\sum_{n=N}^{\infty} a_n^2 \leq \sum_{n=N}^{\infty} \mid a_n \mid}$. By the comparison test, $\displaystyle{\sum_{n=N}^{\infty} a_n^2}$ converges, and so the whole series $\displaystyle{\sum_{n=N}^{\infty} a_n^2}$ converges (absolutely). $\blacksquare$
 Theorem 2: Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ be an absolutely convergent series. Then $\displaystyle{\sum_{n=1}^{\infty} \frac{a_n}{1 + a_n}}$ is an absolutely convergent series.
• Proof: Consider the series $\displaystyle{\sum_{n=1}^{\infty} \frac{\mid a_n \mid}{\mid 1 + a_n \mid}}$. Since $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges absolutely we have $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges and $\lim_{n \to \infty} a_n = 0$. So there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:
(3)
\begin{align} \quad \mid a_n \mid < \frac{1}{2} \end{align}
• But then, for all $n \geq N$ we have that:
(4)
\begin{align} \quad \mid 1 + a_n \mid \geq \mid 1 \mid - \mid a_n \mid > 1 - \frac{1}{2} = \frac{1}{2} \end{align}
• So we have that for all $n \geq N$:
(5)
\begin{align} \quad \frac{\mid a_n \mid}{\mid 1 + a_n \mid} < \frac{\mid a_n \mid}{\frac{1}{2}} = 2 \mid a_n \mid \end{align}
• But this shows that $\displaystyle{\sum_{n=N}^{\infty} \frac{\mid a_n \mid}{\mid 1 + a_n \mid} \leq \sum_{n=N}^{\infty} 2 \mid a_n \mid}$ and so $\displaystyle{\sum_{n=N}^{\infty} \frac{\mid a_n \mid}{\mid 1 + a_n \mid}}$ by the comparison test. So the full series $\displaystyle{\sum_{n=1}^{\infty} \frac{\mid a_n \mid}{\mid 1 + a_n \mid}}$ converges and so $\displaystyle{\sum_{n=1}^{\infty} \frac{a_n}{1 + a_n}}$ converges absolutely. $\blacksquare$