Proof Of The Ratio Test

Let \(\sum^{\infty}_{n=1}a_n\) be a series with nonzero terms. Let \(p\) be defined as:

\[ p = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]

If \(p \gt 1\), then the ratio \(\left| \frac{a_{n+1}}{a_n} \right|\) will eventually be greater than 1; that is, there exists an integer \(N\) such that:

\[ \left| \frac{a_{n+1}}{a_n} \right| \gt 1 \text{ whenever } n \ge N\]

This means \(|a_{n+1}| \gt |a_n|\) whenever \(n \ge N\), so the series \(\sum a_n\) definitely doesn't converge after \(n \ge N\).

If \(p \lt 1\), then choose a number \(r\) such that \(p \lt r \lt 1\):

\[\begin{gather} \left| \frac{a_{n+1}}{a_n} \right| \lt r \text{ whenever } n \ge N \\ | a_{n+1} | \lt |a_n| r \text{ whenever } n \ge N \end{gather}\]

This means:

\[\begin{align} & \ && \ &&| a_{N+1} | &&\lt |a_N| r \\ &| a_{N+2} | &&\lt &&| a_{N+1} |r &&\lt |a_N| r^2 \\ |a_{N+3}| \lt &| a_{N+2} |r &&\lt &&| a_{N+1} |r^3 &&\lt |a_N| r^3 \end{align}\]

In general:

\[|a_{N+k}| \lt |a_N| r^k \text{ for all } k \le 1\]

Consider the geometric series below:

\[ \sum^{\infty}_{k=1} |a_N| r^k = |a_N| r + |a_N| r^2 + |a_N| r^3 + \cdots \]

This is a convergent series since \(0 \lt r \lt 1\). Now consider the series:

\[ \sum^{\infty}_{k=1} |a_{N+k}| = |a_{N+1}| + |a_{N+2}| + |a_{N+3}| + \cdots \]

Since \(|a_{N+k}| \lt |a_N| r^k\) and \(\sum^{\infty}_{k=1} |a_N| r^k\) is convergent, then, by the comparison test, \(\sum^{\infty}_{k=1} |a_{N+k}|\) is also convergent.

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