Ratio test and the Root test [duplicate]

Both the ratio test and the root test define a number (via a limit).

If both limits exist (and shows that the series is convergent), what (if any) is the relation between the 2 numbers ? are they equal ? What is the relation (if any) between them and the original series (other than the fact that they say the series is convergent) ?

Solution 1:

For a non-negative real series $(a_n)_{n \in \mathbb{N}}$, the tests give two (possibly undefined) numbers: let’s call them $L_\textit{root} := \lim_n (a_n)^{\frac{1}{n}}$, and $L_\textit{ratio} := \lim_n \frac{a_{n+1}}{a_n}$.

From Lemma 3 of these notes by Pete L. Clark, it follows that if $L_{\textit{ratio}}$ is defined, then $L_\textit{root}$ is also defined, and they are equal.

This is reasonably intuitive, with a bit of thought: suppose that for $n>N$, the ratio of consecutive terms $\frac{a_{n+1}}{a_n}$ is always close to $L$. Then (still for $n>N$), consider $a_n$ as produced by multiplying $a_N$ by all the later consecutive ratios; so it’s close to $L^{n-N} a_N$, and its $n$th root is close to $(L^{n-N} a_N)^{\frac{1}{n}} = L (\frac{a_N}{L^N})^\frac{1}{n}$. The second factor here, being the $n$th root of a constant, goes to $1$ as $n$ grows; so for sufficiently large $n$, $(a_n)^\frac{1}{n}$ will be close to $L$. (Exercise: make this argument precise — replace each “…close to…” by appropriate specific bounds.)

On the other hand, the converse doesn’t generally hold. $L_\textit{root}$ may be defined even if $L_\textit{ratio}$ is not. For example, set $a_n = 2^n$ when $n$ is even, $a_n = 2^{n-1}$ when $n$ is odd. Then the ratio of consecutive terms alternates between 1 and 4, so $L_\textit{ratio}$ is undefined; but the sequence is close enough to $2^n$ that the root converges, with $L_\textit{root} = 2$.

(Thanks to @David Mitra’s comment for the reference to the linked notes.)

Solution 2:

For your first question:

If both limits exist, they must be equal to each other. In fact, for a sequence of positive terms $(a_n)$, if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty}\root n \of {a_n}$ and moreover, in this case, the two limits are equal to each other. This follows from a more general fact contained in these notes of Pete L. Clark.

I'm not sure if the following answers your second question, but:

In general, there is no relationship between the value of the limit $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ and the value of the sum $\sum\limits_{n=1}^\infty a_n$.
Indeed, here is a silly example showing this:

Suppose $(a_n)$ is a sequence of positive terms and that $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}=r<1$. Then $\sum\limits_{n=1}^\infty a_n$ converges, say to $S\ne 0$. Now let $a>0$ and consider the sequence $(b_n)$ defined by $b_n=a\cdot a_n$. Here we have $\lim\limits_{n\rightarrow\infty} {b_{n+1}\over b_n} =\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}= r$. But, $\sum\limits_{n=1}^\infty b_n=aS$.

So if, $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}=r<1$, the corresponding series could possibly converge to any given positive number. The same remark holds for the limit in the Root test.