Show is convergent iff. , and that
(i) assume is convergent.
Because it is convergent, there is some value alpha such that:
I am not sure if I am on the right track, and if I am, what should I do next?
For the forward implication I would break the proof into two parts.
First show that
I assume that you mean the above if k=0 the sequence is constant and equal to 1.
So suppose not( assume k < 0) and use what you have above to get a contradiction. Hint the sequence will be unbounded.
Then after you know that k > 0 you can make a direct argument to show that it converges to zero.
The reverse <= implication is not bad at all.
Instead of doing "k non-negative implies convergent" and "convergent implies k non-negative", it'll probably be easier to do "k non-negative implies convergent" and "k negative implies divergent".
There are probably nicer ways to do it, but for k non-negative, I'd choose m such that 1/m<k. Then , and since you know as , so must , and so . The case where k is negative follows immediately.
Since this is for an analysis class I would be a bit more explicit.
Since we know that
This implies that for every epsilon greater than 0 there exists an N such that ... so set
But this implies that the sequence is bounded. But since k < 0 we can make
as large as we wish.
Let M > 0 be any real number
Then Pick
Then for any n > n'
so we have the desired contradiction.
Lets look at a concreate example.
Let
and set
Now notice if I set
Now for any
if
This shows that if k is negative the sequced can be made as large as we want. There is nothing special about one million the above is true for any positive M so the sequence diverges to infintity.