If Ratio Test Limit exists, then Root Test Limit exists, and both are equal

The limit for ratio test is \lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|}, while the limit for root test is \lim_{n\to\infty}|a_n|^{1/n}. Something special about these two limits is that if the former exists, the latter also exists and they are equal!


Let \lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|=L. There exists N\in\mathbb{N} such that n\geq N \implies ||\frac{a_{n+1}}{a_n}|-L|<\epsilon.

i.e. L-\epsilon<|\frac{a_{n+1}}{a_n}|<L+\epsilon

For n>N,

|a_n|=\frac{|a_n|}{|a_{n-1}|}\cdot \frac{|a_{n-1}|}{|a_{n-2}|}\cdots \frac{|a_{N+1}|}{|a_N|}\cdot |a_N| < (L+\epsilon)^{n-N}\cdot |a_N|.

Taking nth roots,

|a_n|^{1/n}<(L+\epsilon)^\frac{n-N}{n}\cdot |a_N|^{\frac{1}{n}}

Taking limits,

\lim_{n\to\infty}|a_n|^{\frac{1}{n}}\leq (L+\epsilon)

Since \epsilon is arbitrary, \lim_{n\to\infty}|a_n|^{\frac{1}{n}}\leq L.

Similarly, we can show \lim_{n\to\infty}|a_n|^\frac{1}{n}\geq L.

Thus, \lim_{n\to\infty}|a_n|^\frac{1}{n}=L.

This is considered a rather tricky (though not that difficult) proof, hope it helps whoever is searching for it!

Note that the converse is false, we can see that by considering the “rearranged” geometric series: 1/2,1, 1/8, 1/4, 1/32, … (source: https://www.maa.org/sites/default/files/0025570×33450.di021200.02p0190s.pdf)

where the ratio alternates from 2 to 1/8 and hence does not exist.

However, the root test limit of the first 2n terms is defined:

\begin{aligned}    |a_{2n}|&=\frac{|a_{2n}|}{a_{2n-1}}\cdot \frac{|a_{2n-1}|}{|a_{2n-2}|}\cdot \frac{|a_2|}{|a_1|}\cdot |a_1|\\    &=2\cdot \frac{1}{8}\cdot 2 \cdot \frac{1}{8} \cdots 2 \cdot \frac{1}{2}\\    &=2^n \cdot (\frac{1}{8})^{n-1}\cdot \frac{1}{2}\\    &=(\frac{1}{4})^{n-1}    \end{aligned}

Thus, |a_{2n}|^\frac{1}{2n}=\frac{1}{4}^{\frac{n-1}{2n}}\to \frac{1}{2}.

To learn more about epsilon-delta proofs, check out one of the Recommended Analysis Books for Undergraduates.



About mathtuition88

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