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:

\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:

\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:

\begin{align} \quad \mid a_n \mid < \frac{1}{2} \end{align}

- But then, for all $n \geq N$ we have that:

\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 $n \geq N$:

\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. $\blacksquare$