Proof of the Ratio Test

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

∑ k = 1 ∞ a k {\displaystyle \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 → ∞ | a n + 1 a n | = L , L < r {\displaystyle \lim _{n\rightarrow \infty }{\Big |}{\frac {a_{n+1}}{a_{n}}}{\Big |}=L,\quad 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

| a n + 1 a n | < r , w h e n e v e r   n ≥ N {\displaystyle {\Big |}{\frac {a_{n+1}}{a_{n}}}{\Big |}<r,\quad \mathrm {whenever} \ n\geq 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 , w h e n e v e r   n ≥ N {\displaystyle |a_{n+1}|<|a_{n}|r,\quad \mathrm {whenever} \ n\geq N}

If we let n {\displaystyle n} equal N, N + 1, N + 2 in the previous equation we obtain

| 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 {\displaystyle {\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 , w h e n e v e r   k ≥ 1 {\displaystyle |a_{N+k}|<|a_{N}|r^{k},\quad \mathrm {whenever} \ k\geq 1}

Now the series

∑ k = 1 ∞ | a N | r k = | a N | r + | a N | r 2 + | a N | r 3 + … {\displaystyle \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 {\displaystyle r} is known to satisfy 0 < r < 1. By the Comparison Test, the series

∑ n = N + 1 ∞ | a n | = | a N + 1 | + | a N + 2 | + | a N + 3 | + … {\displaystyle \sum _{n=N+1}^{\infty }|a_{n}|=|a_{N+1}|+|a_{N+2}|+|a_{N+3}|+\ldots }

is convergent, and so our given series

∑ k = 1 ∞ a k {\displaystyle \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

| a n + 1 a n | → L > 1 {\displaystyle {\Big |}{\frac {a_{n+1}}{a_{n}}}{\Big |}\rightarrow L>1}

then the ratio

| a n + 1 a n | {\displaystyle {\Big |}{\frac {a_{n+1}}{a_{n}}}{\Big |}}

will eventually be greater than 1; that is, there exists an integer N such that

| a n + 1 a n | > 1 , f o r   n ≥ N {\displaystyle {\Big |}{\frac {a_{n+1}}{a_{n}}}{\Big |}>1,\quad \mathrm {for} \ n\geq N}

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