Analysis - a liminf inequality

[quasar987]

Homework Statement

I'm asked to show that

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

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 [itex]L=\liminf (|a_n|)^{1/n}[/tex] and suppose M>L. Then for all e'>0, there is an N' such that n>N' ==>$|a_{n+1}/a_n|>M-\epsilon'$. 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 ==> $L-\epsilon<(a_{n_k+1})^{1/n_k+1}<L+\epsilon$ <u>and</u> $|a_{n_k+1}/a_{n_k}|>M-\epsilon'$. But <br /> <br /> $$L-\epsilon<(a_{n_k+1})^{1/n_k+1}<L+\epsilon \Leftrightarrow$$ <br /> <br /> $$(L-\epsilon)^{1/n_k+1}<a_{n_k+1}<(L+\epsilon)^{1/n_k+1}\Leftrightarrow$$ <br /> <br /> $$\frac{(L-\epsilon)^{1/n_k+1}}{a_{n_k}}<\frac{a_{n_k+1}}{a_{n_k}}<\frac{(L+\epsilon)^{1/n_k+1}}{a_{n_k}}\Leftrightarrow$$ <br /> $$\frac{(L-\epsilon)^{1/n_k+1}}{(L+\epsilon)^{1/n_k}}<\frac{a_{n_k+1}}{a_{n_k}}<\frac{(L+\epsilon)^{1/n_k+1}}{(L-\epsilon)^{1/n_k}}$$ <br /> <br /> $$\Rightarrow \frac{a_{n_k+1}}{a_{n_k}}<\left(\frac{L+\epsilon}{L-\epsilon}\right)^{1/n_k}}(L+\epsilon)}$$ <br /> <br /> 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.<br /> <br /> 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).[/itex]

 

[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).

 

[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!