Abel's Test for Convergence of Series of Real Numbers

Abel's Test for Convergence of Series of Real Numbers

Recall from The Partial Summation Formula for Series of Real Numbers page that if $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are both sequences of real numbers then:

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

Furthermore we noted that the series $\displaystyle{\sum_{k=1}^{\infty} a_kb_k}$ converges if $(A_nb_{n+1})_{n=1}^{\infty}$ converges and $\displaystyle{\sum_{k=1}^{\infty} A_k(b_{k+1} - b_k)}$ converges

We will now look at another very important test for convergence of series known as Abel's test which, like Dirichlet's Test for Convergence of Series of Real Numbers, relies on the theorem above.

Theorem 1 (Abel's Test): Let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be two sequences of real numbers. If $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges and if $(b_n)_{n=1}^{\infty}$ is monotonic and convergent then $\displaystyle{\sum_{n=1}^{\infty} a_nb_n}$ converges.
  • Proof: Since $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges we must have that the sequence of partial sums $(A_n)_{n=1}^{\infty}$ converges. Therefore this sequence is bounded and so there exists an $M \in \mathbb{R}$, $M > 0$ such that for all $n \in \mathbb{N}$ we have that:
\begin{align} \quad \mid a_n \mid \leq M \end{align}
  • Furthermore since $(b_n)_{n=1}^{\infty}$ is a convergent sequence we see that $\displaystyle{\lim_{n \to \infty} A_nb_{n+1}}$ converges.
  • Additionally we see that:
\begin{align} \quad \sum_{k=1}^{\infty} \mid A_k(b_{k+1} - b_k) \mid = \sum_{k=1}^{\infty} \mid A_k \mid \mid b_{k+1} - b_k \mid \leq \sum_{k=1}^{\infty} M \mid b_{k+1} - b_k \mid = M \sum_{k=1}^{\infty} \mid b_{k+1} - b_k \mid \end{align}
  • There are two cases to consider. Suppose that $(b_n)_{n=1}^{\infty}$ is a monotonically increasing sequence. Then $\displaystyle{\sum_{k=1}^{\infty} \mid b_{k+1} - b_k \mid = \sum_{k=1}^{\infty} (b_{k+1} - b_k)}$ which converges to $-b_1$.
  • Now suppose that $(b_n)_{n=1}^{\infty}$ is a monotonically decreasing sequence. Then $\displaystyle{\sum_{k=1}^{\infty} \mid b_k - b_{k+1} \mid = \sum_{k=1}^{\infty} (b_{k} - b_{k+1})}$ which converges to $b_1$.
  • In either case we see that the series $\displaystyle{\sum_{k=1}^{\infty} \mid b_{k+1} - b_k \mid }$ converges. Therefore the righthand side of the partial summation formula for series of real numbers converges which implies that $\displaystyle{\sum_{n=1}^{\infty} a_nb_n}$ converges. $\blacksquare$
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