Science:Infinite Series Module/Appendices/Proof of The Ratio Test/Proof of The Ratio Test

We can now provide the proof of the ratio test. Recall the ratio test:

The Ratio Test
To apply the ratio test to a given infinite series $\begin{array}{r}\sum _{k=1}^{\mathrm{\infty }}{a}_k,\end{array}$ we evaluate the limit $\begin{array}{r}\underset{k\to \mathrm{\infty }}{\lim }\left|\frac{a_{k+1}}{a_k}\right|=L.\end{array}$ There are three possibilities: if L < 1, then the series converges if L > 1, then the series diverges if L = 1, then the test is inconclusive

Proof

Our proof will be in two parts:

1. Proof of 1 (if L < 1, then the series converges)
2. Proof of 2 (if L > 1, then the series diverges)

Proof of 1 (if L < 1, then the series converges)

Our aim here is to compare the given series

$\sum _{k=1}^{\mathrm{\infty }}{a}_k$

with a convergent geometric series (we will be using a comparison test).

In this first case, L is less than 1, so we may choose any number r such that L < r < 1. Since

$\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L,L<r$

the ratio |a_n+1/a_n| will eventually be less than r. In other words, there exists an integer N such that

$\left|\frac{a_{n+1}}{a_n}\right|<r,\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{r}n\ge N$

This follows from the formal definition of limit, which not all calculus students have covered. Those students who have not covered a formal definition of limit may wish to consult a calculus textbook on this (any of the textbooks listed on the Recommended Resources page cover it).

We can rearranged our expression to

$|a_{n+1}|<|a_n|r,\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{r}n\ge N$

If we let $n$ equal N, N + 1, N + 2 in the previous equation we obtain

$\begin{array}{rl}|a_{N+1}|& <|a_N|r\\ |a_{N+2}|& <|a_{N+1}|r<|a_N|r^2\\ |a_{N+3}|& <|a_{N+2}|r<|a_N|r^3\\ \end{array}$

and, in general,

$|a_{N+k}|<|a_N|r^k,\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{r}k\ge 1$

Now the series

$\sum _{k=1}^{\mathrm{\infty }}|a_N|r^k=|a_N|r+|a_N|r^2+|a_N|r^3+\dots$

is convergent because it is a geometric series whose common ratio $r$ is known to satisfy 0 < r < 1. By the Comparison Test, the series

$\sum _{n=N+1}^{\mathrm{\infty }}|a_n|=|a_{N+1}|+|a_{N+2}|+|a_{N+3}|+\dots$

is convergent, and so our given series

$\sum _{k=1}^{\mathrm{\infty }}{a}_k$

is also convergent (adding a finite number of finite terms to a convergent series will create another convergent series).

Therefore, our series is absolutely convergent (and therefore convergent).

Proof of 2 (if L > 1, then the series diverges)

Here the Divergence Test implies that the given series diverges. Indeed, if

$\left|\frac{a_{n+1}}{a_n}\right|\to L>1$

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

$\left|\frac{a_{n+1}}{a_n}\right|>1,\mathrm{f}\mathrm{o}\mathrm{r}n\ge N$

For this same $N,$ chaining these inequalities together shows that $|a_{N+k}|>|a_N|>0$ for all $k\ge 1$. This clearly implies that the sequence $\{a_n\}$ cannot converge to 0. Therefore, the given series diverges by the Divergence Test.