# 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{aligned}\sum _{k=1}^{\infty }a_{k},\end{aligned}} we evaluate the limit

{\begin{aligned}\lim _{k\rightarrow \infty }{\Big |}{\frac {a_{k+1}}{a_{k}}}{\Big |}=L.\end{aligned}} There are three possibilities:

1. if L < 1, then the series converges
2. if L > 1, then the series diverges
3. 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

$\sum _{k=1}^{\infty }a_{k}$ 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

$\lim _{n\rightarrow \infty }{\Big |}{\frac {a_{n+1}}{a_{n}}}{\Big |}=L,\quad L the ratio |an+1/an| will eventually be less than r. In other words, there exists an integer N such that

${\Big |}{\frac {a_{n+1}}{a_{n}}}{\Big |} 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

$|a_{n+1}|<|a_{n}|r,\quad \mathrm {whenever} \ n\geq N$ If we let $n$ equal N, N + 1, N + 2 in the previous equation we obtain

{\begin{aligned}|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{aligned}} and, in general,

$|a_{N+k}|<|a_{N}|r^{k},\quad \mathrm {whenever} \ k\geq 1$ Now the series

$\sum _{k=1}^{\infty }|a_{N}|r^{k}=|a_{N}|r+|a_{N}|r^{2}+|a_{N}|r^{3}+\ldots$ 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

$\sum _{n=N+1}^{\infty }|a_{n}|=|a_{N+1}|+|a_{N+2}|+|a_{N+3}|+\ldots$ is convergent, and so our given series

$\sum _{k=1}^{\infty }a_{k}$ 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

${\Big |}{\frac {a_{n+1}}{a_{n}}}{\Big |}\rightarrow L>1$ then the ratio

${\Big |}{\frac {a_{n+1}}{a_{n}}}{\Big |}$ will eventually be greater than 1; that is, there exists an integer N such that

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