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Prove Raabe’s test for the convergence of a series

Consider the series \sum a_n where a_n > 0 for all n. Raabe’s test for convergence states that if there exists an integer N \geq 1 and a constant r > 0 such that

    \[ \frac{a_{n+1}}{a_n} \leq 1 - \frac{1}{n} - \frac{r}{n} \qquad \text{for all } n \geq N \]

then the series \sum a_n is convergent. The series is divergent if

    \[ \frac{a_{n+1}}{a_n} \geq 1 - \frac{1}{n} \qquad \text{for all } n \geq N. \]


Proof. Assume there is an r > 0 and an N \geq 1 such that

    \[ \frac{a_{n+1}}{a_n} \leq 1 - \frac{1}{n} - \frac{r}{n} \qquad \text{for all } n \geq N. \]

From the previous exercise (Section 10.16, Exercise #15) we know that if there exists an integer N and an r >0 such that

    \[ 0 < r \leq b_n - \frac{b_{n+1}}{a_{n+1}}{a_n} \qquad \text{for all } n \geq N \]

then \sum a_n converges. Taking b_n = n, we have

    \begin{align*}  && \frac{a_{n+1}}{a_n} &\leq 1 - \frac{1}{n} -\frac{r}{n} \\[9pt]  \implies && \frac{n(a_{n+1})}{a_n} &\leq n - 1 - r \\[9pt]  \implies && \frac{n(a_{n+1})}{a_n} + 1 - n &\leq -r \\[9pt]  \implies && r &\leq n - n \left( \frac{a_{n+1}}{a_n} \right) - 1 \\[9pt]  \implies && r &\leq n - \left( \frac{na_{n+1}}{a_n} + 1 \right) \\[9pt]  \implies && r &\leq n - \left( \frac{na_{n+1}}{a_n} + \frac{a_{n+1}}{a_n} \right) &(\frac{a_{n+1}}{a_n} \leq 1 \text{ by assumption})\\[9pt]  \implies && r &\leq n - \frac{(n+1)a_{n+1}}{a_n} \\[9pt]  \implies && r &\leq b_n - \frac{b_{n+1}a_{n+1}}{a_n}. \end{align*}

Hence, \sum a_n converges.
Next, if

    \[ \frac{a_{n+1}}{a_n} \geq 1 - \frac{1}{n} \qquad \text{for all } n \geq N, \]

then we have for all n \geq N,

    \[ \frac{n a_{n+1}}{a_n} \leq n - 1 \quad \implies \quad n - 1 - \frac{na_{n+1}}{a_n} \leq 0. \]

Letting b_n = n-1 we then have

    \[ c_n = b_n - \frac{b_{n+1}a_{n+1}}{a_n} \leq 0. \]

Since we know \sum \frac{1}{n-1} diverges, we may apply the previous exercise to conclude that \sum a_n diverges. \qquad \blacksquare

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