Analysis - a liminf inequality

1. Jun 23, 2007

quasar987

1. The problem statement, all variables and given/known data

$$\liminf |a_{n+1}/a_n|\leq\liminf (|a_n|)^{1/n}$$

3. The attempt at a solution

I tried a proof by contradiction in both "directions", but it did not work. For instance, one "direction" would be to let $M=\liminf |a_{n+1}/a_n|$ and $L=\liminf (|a_n|)^{1/n}[/tex] and suppose M>L. Then for all e'>0, there is an N' such that n>N' ==>[itex]|a_{n+1}/a_n|>M-\epsilon'$. Also, let $\{a_{n_k}\}$ be the subsequence such that $(a_{n_k})^{1/n_k}\rightarrow L$. Then, given e>0, there is an N such that k>N ==> $L-\epsilon<(a_{n_k+1})^{1/n_k+1}<L+\epsilon$ and $|a_{n_k+1}/a_{n_k}|>M-\epsilon'$. But

$$L-\epsilon<(a_{n_k+1})^{1/n_k+1}<L+\epsilon \Leftrightarrow$$

$$(L-\epsilon)^{1/n_k+1}<a_{n_k+1}<(L+\epsilon)^{1/n_k+1}\Leftrightarrow$$

$$\frac{(L-\epsilon)^{1/n_k+1}}{a_{n_k}}<\frac{a_{n_k+1}}{a_{n_k}}<\frac{(L+\epsilon)^{1/n_k+1}}{a_{n_k}}\Leftrightarrow$$
$$\frac{(L-\epsilon)^{1/n_k+1}}{(L+\epsilon)^{1/n_k}}<\frac{a_{n_k+1}}{a_{n_k}}<\frac{(L+\epsilon)^{1/n_k+1}}{(L-\epsilon)^{1/n_k}}$$

$$\Rightarrow \frac{a_{n_k+1}}{a_{n_k}}<\left(\frac{L+\epsilon}{L-\epsilon}\right)^{1/n_k}}(L+\epsilon)}$$

And then I aked wheter we could find an e such that this inequality would contradict $|a_{n_k+1}/a_{n_k}|>M-\epsilon'$, but obviously not, since as n_k gets bigger and bigger, the inequality becomes less restrictive rather than the opposite...

I also tried the same thing with starting with the fact that $\{a_{n_k}\}$ was the subsequence that converged to M. I arrived at the same kind of uninteresting inequality.

Using the same ideas, I also tried to contradict the fact that if L is the liminf, then for all N, there is an n>N with (a_n)^(1/n)>L+e (and analogously for M).

Last edited: Jun 23, 2007
2. Jun 27, 2007

quasar987

bump!

This should not be so difficult; it is problem 51 in the book (odd numbered) and there isn't even a hint at the back of the book.

Also, it may be helpful to know (problem 47) that the ratio and root tests can be generalized as follow: "If limsup|a_{n+1}/a_n|<1, the series converges, and if liminf|a_{n+1}/a_n|>1, the series diverges." (and analogously for (a_n)^1/n).

3. Jul 2, 2007

quasar987

I emailed a professor and he emailed me back a nice pdf with the solution. It involves letting b_n=ln|a_n|, so that the inequality to show is equivalent to showing that liminf(b_n - b_{n-1}) < liminf(b_n/n) and then letting c_n = b_n - b_{n-1}, so it's equivalent to showing liminf(c_n)<liminf((c_1+...+c_n)/n). then let L=liminf(c_n), M=liminf((c_1+...+c_n)/n) and plug the fact that for any e>0, there's N s.t. n>N ==> c_n > L - e into (c_1+...+c_n)/n. It will result by taking the limits that $M\geq L - \epsilon$. Since this is true for any e, we get the desired $M\geq L$.

There is a similar inequality for limsup: $$\limsup |a_{n+1}/a_n| \geq \limsup (|a_n|)^{1/n}$$.

This is nice because it implies that if the generalized root test is inconclusive, then so will the generalized ratio test so no need to waste time trying it!

Last edited: Jul 2, 2007