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

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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)

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. $\blacksquare$

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