A Test for Absolute Convergence of a Series

[Euge]

Let $\{a_n\}_{n = 1}^\infty$ be a sequence of real numbers such that for some real number $p > 1$, $\frac{a_n}{a_{n+1}} = 1 + \frac{p}{n} + b_n$ where $\sum b_n$ converges absolutely. Show that $\sum a_n$ also converges absolutely.

 

[julian]

Spoiler

If a sequence divergences to $+ \infty$ then so does every subsequence. For any $r > 0$, we must have that $\frac{r}{n} \leq |b_n|$ for only a finite number of terms, otherwise $\infty = \sum^\infty \frac{r}{n} \leq \sum^\infty |b_n|$ where the sum is taken over an arbitrary subsequence. Therefore, there exists a $N$ such that $\frac{r}{n} > |b_n|$ for all $n > N$. Therefore, there exists a $N$ such that $\frac{r}{n} + b_n > 0$ for all $n > N$.

Define a $q$ such that $1 < q < p$. There exists an $N$ such that $\frac{p-q}{n} + b_n > 0$ for all $n > N$. From $\frac{a_n}{a_{n+1}} = 1 + \frac{p}{n} + b_n$ we have

\begin{align*}
\dfrac{|a_n|}{|a_{n+1}|} & = |1 + \frac{q}{n} + \frac{p-q}{n} + b_n|
\nonumber \\
& > 1 + \frac{q}{n}
\end{align*}

for $n > N$.

Rearranged:

\begin{align*}
n \left( \dfrac{|a_n|}{|a_{n+1}|} - 1 \right) > q \qquad (*)
\end{align*}

The Raabe-Duhamel's test: Let $\{ c_n \}$ be a sequence of positive numbers. Define

\begin{align*}
\rho_n := n \left( \dfrac{c_n}{c_{n+1}} - 1 \right)
\end{align*}

if

\begin{align*}
L = \lim_{n \rightarrow \infty} \rho_n
\end{align*}

exists and $L > 1$ the series converges.

From $(*)$ we have

\begin{align*}
L = \lim_{n \rightarrow \infty} \rho_n = \lim_{n \rightarrow \infty} n \left( \dfrac{|a_n|}{|a_{n+1}|} - 1 \right) > q > 1 .
\end{align*}

Hence, $\sum |a_n|$ converges.