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. $\blacksquare$