1. ## two easy questions

Here's something I thought would be easy to prove (if I'm corret of course)

If $x_n$ -> x and x >a, then $x_n>a$ for infinitly many values of n.

My working: we need $|x_n -x|<|x-a|= x-a$ for n sufficently large

-> $|x_n|<|x|+x-a$

whatever I do I can't seem to deduce x_n > a for all but finite n

My second question is very simple. In the ratio test, if the limit does not exist what does limit superior greater than 1 imply?

Thanks.

2. ## Re: two easy questions

Originally Posted by boromir
Here's something I thought would be easy to prove (if I'm corret of course)
If $x_n$ -> x and x >a, then $x_n>a$ for infinitly many values of n.
Let $\varepsilon = \frac{{x - a}}{2} > 0$

If $|x_n-x|<\varepsilon$ then $a<\frac{a+x}{2}.

3. ## Re: two easy questions

Originally Posted by boromir
My second question is very simple. In the ratio test, if the limit does not exist what does limit superior greater than 1 imply?
If the inequality ${a_{n+1}\over a_n} > 1$ holds for all but finitely many n, then the series diverges. You can't say anything about limsup i think