1

I'm trying to prove that, if $a_n\rightarrow\infty$ and if $b_n\rightarrow\infty$, then $a_nb_n\rightarrow \infty$. Here's my proof: enter image description here

But what happens when $0<K<1,$ in which case $K>K^2$?

Thanks

beep-boop
  • 11,595
  • It doesn't matter. The definition means that if you happen to be given $K<1$ then it will be covered by say, the case $K=1$. Just give it a think; an $N$ that works for some $K$ also works for every $K' < K$. – FireGarden Apr 22 '14 at 22:22
  • Sorry, I don't quite follow. – beep-boop Apr 22 '14 at 22:36
  • Given some $K$, you can find an $N$ such that $a_n>K$ whenever $n>N$. Now if you wanted $a_n>L$ where $L<K$, then of course $n>N$ will still work; we'll have $a_n>K>L$. The same will apply to the product, just with some slightly fiddly details to write. – FireGarden Apr 22 '14 at 22:47
  • Got it. Thanks. Should I explain the last bit in the proof or can I just say that it's trivially true for $K<1$? – beep-boop Apr 22 '14 at 22:57
  • 1
    Well, I guess it's a matter of why you're writing this proof. If it is some form of coursework then I'd write out the details fully (as I always have with any coursework!), but otherwise some form of explanation as why it won't matter would be enough, if the reader could see the detail for themselves from it. – FireGarden Apr 22 '14 at 23:03

1 Answers1

1

Once you prove the statement for large K it is automatically proven for any smaller K: We know eventually the product is bigger than 10000000 so of course it is eventually bigger than 1/2.

If you're writing out a neat proof, for K between 0 and 1, just take N to be whatever it would be for K=1.

mathematician
  • 2,450