1. ## Convergence

Show $\left(\frac{1}{n^k}\right)_{n\in\mathbb{N}$ is convergent iff. $k\geq 0$, and that $\left(\frac{1}{n^k}\right)\rightarrow 0, \ \forall k>0$

(i) assume $\left(\frac{1}{n^k}\right)$ is convergent.
Because it is convergent, there is some value alpha such that:
$\left(\frac{1}{n^k}\right)\rightarrow\alpha$
$\Rightarrow\left |\frac{1}{n^k}-\alpha\right |<\epsilon$

I am not sure if I am on the right track, and if I am, what should I do next?

2. For the forward implication I would break the proof into two parts.

First show that

$k > 0$

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.

3. 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 $\sqrt[m]{n}, and since you know $\sqrt[m]n\to\infty$ as $n\to\infty$, so must $n^k$, and so $\frac1{n^k}\to0$. The case where k is negative follows immediately.

$\left(\frac{1}{n^k}\right)$ is convergent and k < 0.
Since k < 0, we can right $k=-r, \ r\in\mathbb{R}, \ r>0$
$\Rightarrow\left(\frac{1}{n^k}\right)=\left(\frac{ 1}{n^{-r}}\right)=(n^r)$
$\Rightarrow\text{as} \ n\rightarrow\infty \ (n^r)\rightarrow\infty$
Thus, we have reached contradiction and the sequence is convergent for $k\geq 0$

5. (b) $\left(\frac{1}{n^k}\right)\rightarrow 0, \ \forall k>0$
$\left(\frac{1}{n^k}\right) \ \text{and as} \ n\to\infty, \ \left(\frac{1}{\infty^k}\right)=\left(\frac{1}{ \infty }\right)\to 0$

Good?

6. Originally Posted by dwsmith
$\left(\frac{1}{n^k}\right)$ is convergent and k < 0.
Since k < 0, we can right $k=-r, \ r\in\mathbb{R}, \ r>0$
$\Rightarrow\left(\frac{1}{n^k}\right)=\left(\frac{ 1}{n^{-r}}\right)=(n^r)$
$\Rightarrow\text{as} \ n\rightarrow\infty \ (n^r)\rightarrow\infty$
Thus, we have reached contradiction and the sequence is convergent for $k\geq 0$
Since this is for an analysis class I would be a bit more explicit.

Since we know that

$\left\{\frac{1}{n^k} \right\} \to 0$

This implies that for every epsilon greater than 0 there exists an N such that ... so set

$\epsilon=\frac{1}{2} \implies \exists N \in \mathbb{N} \text{ such that }\left|\frac{1}{n^k}-\alpha \right| < \frac{1}{2} \implies -\frac{1}{2}+\alpha < \frac{1}{n^k} < \frac{1}{2}+\alpha$

But this implies that the sequence is bounded. But since k < 0 we can make

$\left\{\frac{1}{n^k} \right\}$

as large as we wish.

Let M > 0 be any real number

Then Pick

$N'=\left( \frac{1}{M}\right)^{\frac{1}{k}}$

Then for any n > n'

$\left\{\frac{1}{n^k} \right\} > M$

so we have the desired contradiction.

7. Originally Posted by TheEmptySet
Let M > 0 be any real number

Then Pick

$N'=\left( \frac{1}{M}\right)^{\frac{1}{k}}$

Then for any n > n'

$\left\{\frac{1}{n^k} \right\} > M$

so we have the desired contradiction.
I don't quite understand what is going on here.

8. Originally Posted by dwsmith
I don't quite understand what is going on here.
Lets look at a concreate example.

Let

$k=-\frac{1}{2}$

and set

$M=1,000,000=10^6$

Now notice if I set

$N'=\left( \frac{1}{10^6} \right)^{ \frac{1}{-\frac{1}{2}}$ $=10^{12}$

Now for any

$n > N'=10^{12}$

$a_{10^{12}}=\frac{1}{(10^{12})^k}=10^{-12k}=10^{6}$

if $k=-\frac{1}{2}$

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.