Convergence of Series of Products of Sequences with the P.S. Formula

# Convergence of Series of Products of Sequences with the Partial Summation Formula

Recall from the The Partial Summation Formula for Series of Real Numbers page that if $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are two sequence of real numbers, then we can consider the series $\displaystyle{\sum_{n=1}^{\infty} a_nb_n}$, that is, the series whose terms are the product of corresponding terms in the sequences $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$. We found a very nice formula for such finite series known as the partial summation formula for series of real numbers. For each $n \in \mathbb{N}$, it is given by:

(1)
\begin{align} \quad \sum_{k=1}^{n} a_kb_k = A_kb_{k+1} - \sum A_k(b_{k+1} - b_k) \end{align}

We saw that the series $\displaystyle{\sum_{n=1}^{\infty} a_nb_n}$ will converge if both the sequence $(A_nb_{n+1})_{n=1}^{\infty}$ and the series $\displaystyle{\sum_{n=1}^{\infty} A_n(b_{n+1} - b_n)}$ converge. We will now look at some nice results applying this formula.

 Theorem 1: Let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be two sequences of real numbers, and let $(A_n)_{n=1}^{\infty}$ denote the sequence of partial sums for the series $\displaystyle{\sum_{n=1}^{\infty} a_n}$. If $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges and $\displaystyle{\sum_{n=1}^{\infty} (b_n - b_{n+1})}$ converges absolutely, then $\displaystyle{\sum_{n=1}^{\infty} a_nb_n}$ converges.
• Proof: We begin by showing that the sequence $(A_nb_{n+1})_{n=1}^{\infty}$ converges. Since $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges, the sequence of partial sums $(A_n)_{n=1}^{\infty}$ converges.
• Also, since $\displaystyle{\sum_{n=1}^{\infty} (b_n - b_{n+1})}$ converges absolutely, $\displaystyle{\sum_{n=1}^{\infty} (b_n - b_{n+1})}$ converges. Let $(B_n)_{n=1}^{\infty}$ denote the sequence of partial sums for this series. Then $\displaystyle{\lim_{n \to \infty} B_n = B}$ for some $B \in \mathbb{R}$. Furthermore, notice that for each $n \in \mathbb{N}$ we have that:
(2)
\begin{align} \quad B_n = \sum_{k=1}^{n} (b_k - b_{k+1}) = (b_1 - b_2) + (b_2 - b_3) + ... + (b_n - b_{n+1}) = b_1 - b_{n+1} \end{align}
• Therefore we see that $\displaystyle{\lim_{n \to \infty} B_n = \lim_{n \to \infty} (b_1 - b_{n+1}) = b_1 - \lim_{n \to \infty} b_{n+1} = B}$. So $\displaystyle{\lim_{n \to \infty} b_{n+1} = b_1 - B}$ which shows that $(b_{n+1})_{n=1}^{\infty}$ converges. The product of two convergent sequences is always a convergent sequence, so $(A_nb_{n+1})_{n=1}^{\infty}$ converges $(*)$
• We now show that the series $\displaystyle{\sum_{n=1}^{\infty} A_n(b_{n+1} - b_n)}$ converges. Since $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges, the sequence of partial sums $(A_n)_{n=1}^{\infty}$ converges. Every convergent sequence of real numbers is bounded, so there exists an $M \in \mathbb{R}$, $M > 0$ such that $\mid A_n \mid \leq M$ for all $n \in \mathbb{N}$. Notice that:
(3)
\begin{align} \quad \sum_{n=1}^{\infty} \mid A_n(b_{n+1} - b_n) \mid = \sum_{n=1}^{\infty} \mid A_n \mid \mid b_{n+1} - b_n \mid \leq \sum_{n=1}^{\infty} M \mid b_{n+1} - b_n \mid = M \sum_{n=1}^{\infty} \mid b_{n+1} - b_n \mid \end{align}
• But $\displaystyle{\sum_{n=1}^{\infty} \mid b_{n+1} - b_n \mid}$ converges since $\displaystyle{\sum_{n=1}^{\infty} (b_n - b_{n+1})}$ converges absolutely. So $\displaystyle{\sum_{n=1}^{\infty} \mid A_n(b_{n+1} - b_n) \mid}$ converges which shows that $\displaystyle{\sum_{n=1}^{\infty} A_n(b_{n+1} - b_n)}$ converges absolutely and hence converges $(**)$.
• From $(*)$ and $(**)$ we can use the partial summation formula for series of real numbers to conclude that $\displaystyle{\sum_{n=1}^{\infty} a_nb_n}$ converges. $\blacksquare$
 Theorem 2: Let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be two sequences of real numbers, and let $(A_n)_{n=1}^{\infty}$ denote the sequence of partial sums for the series $\displaystyle{\sum_{n=1}^{\infty} a_n}$. If $(A_n)_{n=1}^{\infty}$ is bounded, $\displaystyle{\sum_{n=1}^{\infty} (b_n - b_{n+1})}$ converges absolutely, and $\displaystyle{\lim_{n \to \infty} b_n = 0}$, then $\displaystyle{\sum_{n=1}^{\infty} a_nb_n}$ converges.
• Proof: We begin by establishing the convergence of the sequence $(A_nb_{n+1})_{n=1}^{\infty}$. Since $(A_n)_{n=1}^{\infty}$ is bounded and $(b_{n+1})_{n=1}^{\infty}$ converges to $0$, this implies that $(A_nb_{n+1})_{n=1}^{\infty}$ converges to $0$, i.e., $(A_nb_{n+1})_{n=1}^{\infty}$ converges. $(*)$
• Now Furthermore, since $(A_n)_{n=1}^{\infty}$ has is bounded, there exists an $M \in \mathbb{R}$, $M > 0$ such that for all $n \in \mathbb{N}$ we have that $\mid A_n \mid \leq M$. So then:
(4)
\begin{align} \quad \sum_{n=1}^{\infty} \mid A_n(b_{n+1} - b_n) \mid = \sum_{n=1}^{\infty} \mid A_n \mid \mid b_{n+1} - b_n \mid \leq \sum_{n=1}^{\infty} M \mid b_{n+1} - b_n \mid = M \sum_{n=1}^{\infty} \mid b_{n+1} - b_n \mid \end{align}
• So, as before, we are given that $\displaystyle{\sum_{n=1}^{\infty} \mid b_{n+1} - b_n \mid }$ converges and so by the comparison test, $\displaystyle{\sum_{n=1}^{\infty} \mid A_n(b_{n+1} - b_n) \mid}$ converges. So $\displaystyle{\sum_{n=1}^{\infty} A_n(b_{n+1} - b_n)}$ converges absolutely, i.e., converges. $(**)$
• From $(*)$ and $(**)$ we conclude from using the partial summation formula that $\displaystyle{\sum_{n=1}^{\infty} a_nb_n}$ converges. $\blacksquare$